right triangles and trigonometry homework 6

Right Triangle Trigonometry Calculator

Table of contents

The right triangle trigonometry calculator can help you with problems where angles and triangles meet: keep reading to find out:

• The basics of trigonometry;
• How to calculate a right triangle with trigonometry;
• A worked example of how to use trigonometry to calculate a right triangle with steps;

And much more!

Basics of trigonometry

Trigonometry is a branch of mathematics that relates angles to the length of specific segments . We identify multiple trigonometric functions: sine, cosine, and tangent, for example. They all take an angle as their argument, returning the measure of a length associated with the angle itself. Using a trigonometric circle , we can identify some of the trigonometric functions and their relationship with angles.

As you can see from the picture, sine and cosine equal the projection of the radius on the axis, while the tangent lies outside the circle. If you look closely, you can identify a right triangle using the elements we introduced above: let's discover the relationship between trigonometric functions and this shape.

Right triangles trigonometry calculations

Consider an acute angle in the trigonometric circle above: notice how you can build a right triangle where:

• The radius is the hypotenuse; and
• The sine and cosine are the catheti of the triangle.

α \alpha α is one of the acute angles, while the right angle lies at the intersection of the catheti (sine and cosine)

Let this sink in for a moment: the length of the cathetus opposite from the angle α \alpha α is its sine , sin ⁡ ( α ) \sin(\alpha) sin ( α ) ! You just found an easy and quick way to calculate the angles and sides of a right triangle using trigonometry.

The complete relationships between angles and sides of a right triangle need to contain a scaling factor, usually the radius (the hypotenuse). Identify the opposite and adjacent . We can then write:

By switching the roles of the legs, you can find the values of the trigonometric functions for the other angle.

Taking the inverse of the trigonometric functions , you can find the values of the acute angles in any right triangle.

Using the three equations above and a combination of sides, angles, or other quantities, you can solve any right triangle . The cases we implemented in our calculator are:

• Solving the triangle knowing two sides ;
• Solving the triangle knowing one angle and one side ; and
• Solving the triangle knowing the area and one side .

Example of right triangle trigonometry calculations with steps

Take a right triangle with hypotenuse c = 5 c = 5 c = 5 and an angle α = 38 ° \alpha=38\degree α = 38° . Surprisingly enough, this is enough data to fully solve the right triangle! Follow these steps:

• Calculate the third angle: β = 90 ° − α \beta = 90\degree - \alpha β = 90° − α .
• sin ⁡ ( α ) = 0.61567 \sin(\alpha) = 0.61567 sin ( α ) = 0.61567 .
• o p p o s i t e = sin ⁡ ( α ) ⋅ h y p o t e n u s e = 0.61567 ⋅ 5 = 3.078 \mathrm{opposite} = \sin(\alpha)\cdot\mathrm{hypotenuse} = 0.61567 \cdot 5 = 3.078 opposite = sin ( α ) ⋅ hypotenuse = 0.61567 ⋅ 5 = 3.078 .
• a d j a c e n t = 0.788 ⋅ 5 = 3.94 \mathrm{adjacent} = 0.788\cdot 5 = 3.94 adjacent = 0.788 ⋅ 5 = 3.94 .

More trigonometry and right triangles calculators (and not only)

If you liked our right triangle trigonometry calculator, why not try our other related tools? Here they are:

• The trigonometry calculator ;
• The cosine triangle calculator ;
• The sine triangle calculator ;
• The trig triangle calculator ;
• The trig calculator ;
• The sine cosine tangent calculator ;
• The tangent ratio calculator ; and
• The tangent angle calculator .

How do I apply trigonometry to a right triangle?

To apply trigonometry to a right triangle, remember that sine and cosine correspond to the legs of a right triangle . To solve a right triangle using trigonometry:

• sin(α) = opposite/hypotenuse ; and
• cos(α) = adjacent/hypotenuse .
• By taking the inverse trigonometric functions , we can find the value of the angle α .
• You can repeat the procedure for the other angle.

What is the hypotenuse of a triangle with α = 30° and opposite leg a = 3?

The length of the hypotenuse is 6 . To find this result:

• Calculate the sine of α : sin(α) = sin(30°) = 1/2 .
• Apply the following formula: sin(α) = opposite/hypotenuse hypotenuse = opposite/sin(α) = 3 · 2 = 6 .

Can I apply right-triangle trigonometric rules in a non-right triangle?

Not directly: to apply the relationships between trigonometric functions and sides of a triangle, divide the shape alongside one of the heights lying inside it. This way, you can split the triangle into two right triangles and, with the right combination of data, solve it!

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Chapter 4: Trigonometric Functions

Section 4.3: right triangle trigonometry, learning outcomes.

• Use right triangles to evaluate trigonometric functions.
• Find function values for 30° (π/6),   45° (π/4), and 60° (π/3).
• Use cofunctions of complementary angles.
• Use the definitions of trigonometric functions of any angle.
• Use right triangle trigonometry to solve applied problems.

Using Right Triangles to Evaluate Trigonometric Functions

In earlier sections, we used a unit circle to define the trigonometric functions . In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of [latex]t[/latex] is its value at [latex]t[/latex] radians. First, we need to create our right triangle. Figure 1 shows a point on a unit circle of radius 1. If we drop a vertical line segment from the point [latex]\left(x,y\right)\\[/latex] to the x -axis, we have a right triangle whose vertical side has length [latex]y[/latex] and whose horizontal side has length [latex]x[/latex]. We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.

Likewise, we know

These ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using [latex]\left(x,y\right)[/latex] coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of [latex]x[/latex], we will call the side between the given angle and the right angle the adjacent side to angle [latex]t[/latex]. (Adjacent means “next to.”) Instead of [latex]y[/latex], we will call the side most distant from the given angle the opposite side from angle [latex]t[/latex]. And instead of [latex]1[/latex], we will call the side of a right triangle opposite the right angle the hypotenuse . These sides are labeled in Figure 2.

Figure 2.  The sides of a right triangle in relation to angle [latex]t[/latex].

Understanding Right Triangle Relationships

Given a right triangle with an acute angle of [latex]t[/latex],

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “ S ine is o pposite over h ypotenuse, C osine is a djacent over h ypotenuse, T angent is o pposite over a djacent.”

How To: Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle.

• Find the sine as the ratio of the opposite side to the hypotenuse
• Find the cosine as the ratio of the adjacent side to the hypotenuse.
• Find the tangent is the ratio of the opposite side to the adjacent side.

Example 1: Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure 3, find the value of [latex]\cos \alpha[/latex].

The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so:

[latex]\begin{align}\cos \left(\alpha \right)=\frac{\text{adjacent}}{\text{hypotenuse}} =\frac{15}{17} \end{align}[/latex]

Given the triangle shown in Figure 4, find the value of [latex]\text{sin}t[/latex].

[latex]\frac{7}{25}[/latex]

Relating Angles and Their Functions

When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure 5. The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

Figure 5. The side adjacent to one angle is opposite the other.

We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.

How To: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.

• If needed, draw the right triangle and label the angle provided.
• Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
• sine as the ratio of the opposite side to the hypotenuse
• cosine as the ratio of the adjacent side to the hypotenuse
• tangent as the ratio of the opposite side to the adjacent side
• secant as the ratio of the hypotenuse to the adjacent side
• cosecant as the ratio of the hypotenuse to the opposite side
• cotangent as the ratio of the adjacent side to the opposite side

Example 2: Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure 6, evaluate [latex]\sin \alpha[/latex], [latex]\cos \alpha[/latex], [latex]\tan \alpha[/latex], [latex]\sec \alpha[/latex], [latex]\csc \alpha [/latex], and [latex]\cot \alpha[/latex].

[latex]\begin{align}&\sin \alpha =\frac{\text{opposite }\alpha }{\text{hypotenuse}}=\frac{4}{5}\\ &\cos \alpha =\frac{\text{adjacent to }\alpha }{\text{hypotenuse}}=\frac{3}{5} \\ &\tan \alpha =\frac{\text{opposite }\alpha }{\text{adjacent to }\alpha }=\frac{4}{3} \\ &\sec \alpha =\frac{\text{hypotenuse}}{\text{adjacent to }\alpha }=\frac{5}{3} \\ &\csc \alpha =\frac{\text{hypotenuse}}{\text{opposite }\alpha }=\frac{5}{4} \\ &\cot \alpha =\frac{\text{adjacent to }\alpha }{\text{opposite }\alpha }=\frac{3}{4} \end{align}[/latex]

Using the triangle shown in Figure 7, evaluate [latex]\sin t[/latex], [latex]\cos t[/latex], [latex]\tan t[/latex], [latex]\sec t[/latex], [latex]\csc t[/latex], and [latex]\cot t[/latex].

[latex]\begin{align}&\sin t=\frac{33}{65},\cos t=\frac{56}{65},\tan t=\frac{33}{56}, \\ &\sec t=\frac{65}{56},\csc t=\frac{65}{33},\cot t=\frac{56}{33} \end{align}[/latex]

Finding Trigonometric Functions of Special Angles Using Side Lengths

We have already discussed the trigonometric functions as they relate to the special angles on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of [latex]30^\circ [/latex], [latex]60^\circ [/latex], and [latex]45^\circ[/latex], however, remember that when dealing with right triangles, we are limited to angles between [latex]0^\circ \text{ and } 90^\circ[/latex].

Suppose we have a [latex]30^\circ ,60^\circ ,90^\circ [/latex] triangle, which can also be described as a [latex]\frac{\pi }{6}, \frac{\pi }{3},\frac{\pi }{2}[/latex] triangle. The sides have lengths in the relation [latex]s,\sqrt{3}s,2s[/latex]. The sides of a [latex]45^\circ ,45^\circ ,90^\circ [/latex] triangle, which can also be described as a [latex]\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2}[/latex] triangle, have lengths in the relation [latex]s,s,\sqrt{2}s[/latex]. These relations are shown in Figure 8.

Figure 8. Side lengths of special triangles

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

How To: Given trigonometric functions of a special angle, evaluate using side lengths.

• Use the side lengths shown in Figure 8 for the special angle you wish to evaluate.
• Use the ratio of side lengths appropriate to the function you wish to evaluate.

Example 3: Evaluating Trigonometric Functions of Special Angles Using Side Lengths

Find the exact value of the trigonometric functions of [latex]\frac{\pi }{3}[/latex], using side lengths.

[latex]\begin{align}&\sin \left(\frac{\pi }{3}\right)=\frac{\text{opp}}{\text{hyp}}=\frac{\sqrt{3}s}{2s}=\frac{\sqrt{3}}{2}\\ &\cos \left(\frac{\pi }{3}\right)=\frac{\text{adj}}{\text{hyp}}=\frac{s}{2s}=\frac{1}{2}\\ &\tan \left(\frac{\pi }{3}\right)=\frac{\text{opp}}{\text{adj}}=\frac{\sqrt{3}s}{s}=\sqrt{3}\\ &\sec \left(\frac{\pi }{3}\right)=\frac{\text{hyp}}{\text{adj}}=\frac{2s}{s}=2\\ &\csc \left(\frac{\pi }{3}\right)=\frac{\text{hyp}}{\text{opp}}=\frac{2s}{\sqrt{3}s}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3} \\ &\cot \left(\frac{\pi }{3}\right)=\frac{\text{adj}}{\text{opp}}=\frac{s}{\sqrt{3}s}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} \end{align}[/latex]

Find the exact value of the trigonometric functions of [latex]\frac{\pi }{4}[/latex], using side lengths.

[latex]\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2},\cos \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2},\tan \left(\frac{\pi }{4}\right)=1[/latex], [latex]\sec \left(\frac{\pi }{4}\right)=\sqrt{2},\csc\left(\frac{\pi }{4}\right)=\sqrt{2},\cot \left(\frac{\pi }{4}\right)=1[/latex]

By looking at all six trig functions and their special angles, we can organize it all in one place.

Now that we have this table, we can use it to find the exact values of trigonometric expressions.

Example 4: Evaluating Trigonometric Functions of Special Angles

Find the exact value of the trigonometric function [latex]\frac{\sin\left(\frac{\pi}{6}\right)}{1+\cos^{2}\left(\frac{\pi}{6}\right)}[/latex] using the table above.

[latex]\begin{gathered} & \frac{\frac{1}{2}}{1+\left(\frac{\sqrt{3}}{2}\right)^{2}} && \text{Put in the values from the table. Note the squaring of cosine.} \\ &\frac{\frac{1}{2}}{1+\frac{3}{4}} && \text{Square the fraction.} \\ &\frac{\frac{1}{2}}{\frac{7}{4}} && \text{Get common denominators.} \\ &\left(\frac{1}{2}\right)\left({\frac{4}{7}}\right) && \text{Get common denominators.} \\ &\frac{2}{7} && \text{Simplify}\end{gathered}[/latex]

Find the exact value of the trigonometric function [latex]\sec^{2}\left(45^\circ\right)+\tan^{2}\left(60^\circ\right)[/latex] using the table above.

Using Equal Cofunction of Complements

If we look at the table above, we will notice a pattern. In a right triangle with angles of [latex]\frac{\pi }{6}[/latex] and [latex]\frac{\pi }{3}[/latex], we see that the sine of [latex]\frac{\pi }{3}[/latex], namely [latex]\frac{\sqrt{3}}{2}[/latex], is also the cosine of [latex]\frac{\pi }{6}[/latex], while the sine of [latex]\frac{\pi }{6}[/latex], namely [latex]\frac{1}{2}[/latex], is also the cosine of [latex]\frac{\pi }{3}[/latex].

Figure 9.  The sine of [latex]\frac{\pi }{3}[/latex] equals the cosine of [latex]\frac{\pi }{6}[/latex] and vice versa.

This result should not be surprising because, as we see from Figure 9, the side opposite the angle of [latex]\frac{\pi }{3}[/latex] is also the side adjacent to [latex]\frac{\pi }{6}[/latex], so [latex]\sin \left(\frac{\pi }{3}\right)[/latex] and [latex]\cos \left(\frac{\pi }{6}\right)[/latex] are exactly the same ratio of the same two sides, [latex]\sqrt{3}s[/latex] and [latex]2s[/latex]. Similarly, [latex]\cos \left(\frac{\pi }{3}\right)[/latex] and [latex]\sin \left(\frac{\pi }{6}\right)[/latex] are also the same ratio using the same two sides, [latex]s[/latex] and [latex]2s[/latex].

The interrelationship between the sines and cosines of [latex]\frac{\pi }{6}[/latex] and [latex]\frac{\pi }{3}[/latex] also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to [latex]\pi [/latex], and the right angle is [latex]\frac{\pi }{2}[/latex], the remaining two angles must also add up to [latex]\frac{\pi }{2}[/latex]. That means that a right triangle can be formed with any two angles that add to [latex]\frac{\pi }{2}[/latex] —in other words, any two complementary angles. So we may state a cofunction identity : If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure 10.

Figure 10.  Cofunction identity of sine and cosine of complementary angles

Using this identity, we can state without calculating, for instance, that the sine of [latex]\frac{\pi }{12}[/latex] equals the cosine of [latex]\frac{5\pi }{12}[/latex], and that the sine of [latex]\frac{5\pi }{12}[/latex] equals the cosine of [latex]\frac{\pi }{12}[/latex]. We can also state that if, for a certain angle [latex]t[/latex], [latex]\cos \text{ }t=\frac{5}{13}[/latex], then [latex]\sin \left(\frac{\pi }{2}-t\right)=\frac{5}{13}[/latex] as well.

A General Note: Cofunction Identities

The cofunction identities in radians are listed in the table below.

How To: Given the sine and cosine of an angle, find the sine or cosine of its complement.

• To find the sine of the complementary angle, find the cosine of the original angle.
• To find the cosine of the complementary angle, find the sine of the original angle.

Example 5: Using Cofunction Identities

Write the following as an equivalent cosine expression: [latex]\sin \left(\frac{5\pi}{12}\right)[/latex].

According to the cofunction identities for sine and cosine,

[latex]\sin t=\cos \left(\frac{\pi }{2}-t\right)[/latex].

[latex]\cos \left(\frac{\pi }{2}-\frac{5\pi}{12}\right)=\cos\left(\frac{\pi}{12}\right)[/latex].

Example 6: Using Cofunction Identities

Write the following as an equivalent tangent expression: [latex]\cot\left(25^\circ\right)[/latex].

According to the cofunction identities for tangent and cotangent,

[latex]\tan t=\cot \left(90^\circ-t\right)[/latex].

[latex]\tan \left(90^\circ-25^\circ\right)=\tan\left(65^\circ\right)[/latex].

Write the following as an equivalent cosecant expression: [latex]\sec\left(68^\circ\right)[/latex].

[latex]\csc\left(22^\circ\right)[/latex]

Using Trigonometric Functions

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

How To: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides.

• For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
• Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
• Using the value of the trigonometric function and the known side length, solve for the missing side length.

Example 7: Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure 11.

We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

[latex]\tan \left(30^\circ \right)=\frac{7}{a}[/latex]

We rearrange to solve for [latex]a[/latex].

[latex]\begin{align}a&=\frac{7}{\tan \left(30^\circ \right)} \\ &=12.1\end{align}[/latex]

We can use the sine to find the hypotenuse.

[latex]\sin \left(30^\circ \right)=\frac{7}{c}[/latex]

Again, we rearrange to solve for [latex]c[/latex].

[latex]\begin{align}c&=\frac{7}{\sin \left(30^\circ \right)} \\ &=14 \end{align}[/latex]

A right triangle has one angle of [latex]\frac{\pi }{3}[/latex] and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

[latex]\text{adjacent}=10[/latex]; [latex]\text{opposite}=10\sqrt{3}[/latex] ; missing angle is [latex]\frac{\pi }{6}[/latex]

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer’s eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer’s eye.  In the figure below, [latex]\alpha[/latex] represents the angle of elevation and [latex]\beta[/latex] represents the angle of depression .

How To: Given a tall object, measure its height indirectly.

• Make a sketch of the problem situation to keep track of known and unknown information.
• Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
• At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
• Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
• Solve the equation for the unknown height.

Example 8: Measuring a Distance Indirectly

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures the angle of elevation to be [latex]57^\circ [/latex], as shown in Figure 13. Find the height of the tree.

We know that the angle of elevation is [latex]57^\circ [/latex] and the adjacent side is 30 ft long. The opposite side is the unknown height.

The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of [latex]57^\circ [/latex], letting [latex]h[/latex] be the unknown height.

[latex]\begin{align}&\tan \theta =\frac{\text{opposite}}{\text{adjacent}} \\ &\tan\left(57^\circ \right)=\frac{h}{30}&& \text{Solve for }h. \\ &h=30\tan \left(57^\circ \right)&& \text{Multiply}.\\ &h\approx 46.2&& \text{Use a calculator}. \end{align}[/latex]

The tree is approximately 46 feet tall.

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building with an angle of elevation [latex]\frac{5\pi }{12}[/latex] ? Round to the nearest foot.

About 52 ft

Inverse Trigonometric Function on a calculator

If two sides of a right triangle are given, an inverse trigonometric function can be used to find an acute angle in a triangle. There is an inverse trig button on a calculator that appears above the normal sin, cos, tan buttons on a calculator. A shift or second key is required to select these, and they are [latex]\sin^{-1},\cos^{-1},\tan^{-1}[/latex]. The calculator will return an angle in either radians or degrees, depending on what mode your calculator is in.  We will need the inverse tangent function for bearing.

Now we will will look at some application problems that involve bearings and right triangles.

Example 9: Find the bearing

A semi travels east for 8 miles, makes a right turn, and then travels south for 11 more miles. What is the bearing from the starting point to the semi’s current location?

First we need to draw a picture of the problem:

We want to find the angle [latex]\alpha[/latex] that is inside the triangle. To do this, we will set up a trig definition that relates the two given sides. We will choose tangent. Then we will take the inverse tangent to find the angle: [latex]\begin{align} &\tan\left(\alpha\right)=\frac{11}{8} \\& \tan^{-1}\left(\tan\left(\alpha\right)\right)=\tan^{-1}\left(\frac{11}{8}\right) \\ &\alpha=\tan^{-1}\left(\frac{11}{8}\right) \\& \alpha\approx 54^\circ \end{align}[/latex]

The bearing must be measured from south since the triangle is drawn in the fourth quadrant. On the picture, we must find [latex]\beta[/latex]. Since [latex]\alpha+\beta=90^\circ[/latex], then [latex]\beta=90^\circ-\alpha=36^\circ[/latex]. Therefore our bearing is [latex]S 36^\circ E[/latex].

A cyclist travels west for 7 miles, makes a right turn and travels north for 4 miles. What is the bearing from the starting point to the cyclist’s current position? Round your angle to one decimal place.

[latex]N 60.3^\circ W[/latex]

Example 10: Find the bearing

A jeep travels on a bearing of [latex]N 40^\circ W[/latex] for 6 miles. How far north and how far west is the jeep from its starting point? Round your answers to one decimal place.

We need to find the angle [latex]\alpha[/latex] inside of the triangle before we set up a trig equation: [latex]\alpha=90^\circ-40^\circ=50^\circ[/latex].

Next we will set up a trig function to find the indicated sides of the triangle:

First we will find the number of miles traveled North:

[latex]\begin{align} & \sin\left(50^\circ\right)=\frac{N}{6} \\& 6\sin\left(50^\circ\right)=N \\ & N\approx 4.6\text{ miles} \end{align}[/latex]

Now we will find the number of miles traveled west:

[latex]\begin{align} & \cos\left(50^\circ\right)=\frac{W}{6} \\& 6\cos\left(50^\circ\right)=W \\ & W\approx 3.9\text{ miles} \end{align}[/latex]

A jogger travels on a bearing of [latex]S 53.13^\circ E[/latex] for 5 miles. How far east and how far south is the jogger from their starting point? Round your answers to the nearest mile.

4 miles east, 3 miles south

Convert Percent Grade Into Degrees

Percent grade is often seen on roadways or trails, and it is a measure of steepness.  For example, if a road has a 6% grade, this means the road rises 6 feet over a horizontal distance (run) of 100 feet.  Percent grade is found by dividing the rise over the run as shown in Figure 14. The rise and run are also the opposite and adjacent sides.  We can find [latex]\theta[/latex] by taking the inverse tangent.

In order to calculate the degree measurement of a percent grade, first change the percent into a decimal by dividing by 100. Then take the inverse tangent of this decimal, and this will give the angle in degrees.  This is actually the angle of elevation , which we studied earlier in this section!

How To: Convert Percent Grade inTo Degrees

Make sure your calculator is in degree mode. Given the percent grade, the angle [latex]\theta[/latex] can be calculated using:

[latex]\theta=\tan^{-1}\left(\dfrac{\text{percent grade}}{100}\right)[/latex]

Example 11: Percent Grade to Degrees

Filbert Street, the steepest street in San Francisco, has a 31.5% grade.  What is the angle the street makes with a horizontal line, rounded to one decimal place?

[latex]\theta=\tan^{-1}\left(\dfrac{31.5}{100}\right)\approx 17.5^\circ[/latex]

A road has a grade of 10.9%.  What is the angle the road makes with a horizontal line, rounded to one decimal place?

[latex]\text{About }6.2^\circ[/latex]

Key Equations

Key Concepts

• We can define trigonometric functions as ratios of the side lengths of a right triangle.
• The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle.
• We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur.
• Any two complementary angles could be the two acute angles of a right triangle.
• If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa.
• We can use trigonometric functions of an angle to find unknown side lengths.
• Select the trigonometric function representing the ratio of the unknown side to the known side.
• Right-triangle trigonometry permits the measurement of inaccessible heights and distances.
• The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known.
• Bearings are measured from either from N or S, depending on the first letter of the bearing.
• Percent grade can be expressed as a degree, which is the angle of elevation.

Section 4.3 Homework Exercises

2. When a right triangle with a hypotenuse of 1 is placed in the unit circle, which sides of the triangle correspond to the x- and y-coordinates?

3. The tangent of an angle compares which sides of the right triangle?

4. What is the relationship between the two acute angles in a right triangle?

5. Explain the cofunction identity.

For the following exercises, evaluate the expression.  Rationalize the denominator if necessary.

6. [latex]\sin\frac{\pi}{4}\cos\frac{\pi}{4}-\tan\frac{\pi}{3}[/latex]

7. [latex]\sin\frac{\pi}{4}\cos\frac{\pi}{3}-\tan\frac{\pi}{4}[/latex]

8. [latex]2\sec^{2}\frac{\pi}{4}+\cot^{2}\frac{\pi}{3}[/latex]

9. [latex]3\csc^{2}\frac{\pi}{3}+\cot^{2}\frac{\pi}{4}[/latex]

For the following exercises, use cofunctions of complementary angles.

10. [latex]\cos \left(34^\circ\right)=\sin \left(\text{__}^\circ\right)[/latex]

11. [latex]\cos \left(\frac{\pi }{3}\right)=\sin \text{(___)}[/latex]

12. [latex]\csc \left(21^\circ\right)=\sec \left(\text{___}^\circ \right)[/latex]

13. [latex]\tan \left(\frac{\pi }{4}\right)=\cot \left(\text{__}\right)[/latex]

For the following exercises, find the lengths of the missing sides if side [latex]a[/latex] is opposite angle [latex]A[/latex], side [latex]b[/latex] is opposite angle [latex]B[/latex], and side [latex]c[/latex] is the hypotenuse.

14. [latex]\cos B=\frac{4}{5},a=10[/latex]

15. [latex]\sin B=\frac{1}{2}, a=20[/latex]

16. [latex]\tan A=\frac{5}{12},b=6[/latex]

17. [latex]\tan A=100,b=100[/latex]

18. [latex]\sin B=\frac{1}{\sqrt{3}}, a=2[/latex]

19. [latex]a=5,\measuredangle A={60}^{\circ }[/latex]

20. [latex]c=12,\measuredangle A={45}^{\circ }[/latex]

For the following exercises, use Figure 14 to evaluate each trigonometric function of angle [latex]A[/latex].

21. [latex]\sin A[/latex]

22. [latex]\cos A[/latex]

23. [latex]\tan A[/latex]

24. [latex]\csc A[/latex]

25. [latex]\sec A[/latex]

26. [latex]\cot A[/latex]

For the following exercises, use Figure 15 to evaluate each trigonometric function of angle [latex]A[/latex].

27. [latex]\sin A[/latex]

28. [latex]\cos A[/latex]

29. [latex]\tan A[/latex]

30. [latex]\csc A[/latex]

31. [latex]\sec A[/latex]

32. [latex]\cot A[/latex]

For the following exercises, solve for the unknown sides of the given triangle.

For the following exercises, use a calculator to find the length of each side to four decimal places.

41. [latex]b=15,\measuredangle B={15}^{\circ }[/latex]

42. [latex]c=200,\measuredangle B={5}^{\circ }[/latex]

43. [latex]c=50,\measuredangle B={21}^{\circ }[/latex]

44. [latex]a=30,\measuredangle A={27}^{\circ }[/latex]

45. [latex]b=3.5,\measuredangle A={78}^{\circ }[/latex]

50. A radio tower is located 400 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is [latex]36^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]23^\circ [/latex]. How tall is the tower?

51. A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is [latex]43^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]31^\circ [/latex]. How tall is the tower?

52. A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is [latex]15^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]2^\circ [/latex]. How far is the person from the monument?

53. A 400-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is [latex]18^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]3^\circ [/latex]. How far is the person from the monument?

54. There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be [latex]40^\circ [/latex]. From the same location, the angle of elevation to the top of the antenna is measured to be [latex]43^\circ [/latex]. Find the height of the antenna.

55. There is lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be [latex]36^\circ [/latex]. From the same location, the angle of elevation to the top of the lightning rod is measured to be [latex]38^\circ [/latex]. Find the height of the lightning rod.

56. A 33-ft ladder leans against a building so that the angle between the ground and the ladder is [latex]80^\circ [/latex]. How high does the ladder reach up the side of the building?

57. A 23-ft ladder leans against a building so that the angle between the ground and the ladder is [latex]80^\circ [/latex]. How high does the ladder reach up the side of the building?

58. The angle of elevation to the top of a building in New York is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building.

59. The angle of elevation to the top of a building in Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building.

60. Assuming that a 370-foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be [latex]60^\circ [/latex], how far from the base of the tree am I?

61. A car travels west for 5 miles, turns left, and then travels south for 9 miles.  What is the bearing from car’s starting position to its current position?  Round your answer to two decimal places.

62. A truck travels east for 4 miles, turns left, and then travels north for 6 miles.  What is the bearing from the truck’s starting position to its current position? Round your answer to two decimal places.

63. An ant travels on a bearing of [latex]N 22^\circ E[/latex] for 36 inches.  How far east and how far north is the ant from its starting position?  Round your answers to two decimal places.

64. A spider crawls on a bearing of [latex]S 34^\circ W[/latex] for 20 inches.  How far west and how far south is the spider from its starting point? Round your answers to two decimal places.

65. Originally built in 1901 by Colonel J.W. Eddy, Angels Flight in Los Angeles is said to be the world’s shortest incorporated railway.  The counterbalanced cars, controlled by cables, travel a 33% grade for 315 feet.  What is the angle the track makes with a horizontal line, rounded to one decimal place?

66. The Saluda Grade is the steepest standard-gauge mainline railway grade in the United States.  Between Melrose and Saluda, North Carolina, the maximum grade is 4.9% for about 300 feet.  What is the angle the track makes with a horizontal line, rounded to one decimal place?

• Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution
• Introduction to Trigonometric Functions Using Triangles. Authored by : Mathispower4u. Located at : https://youtu.be/Ujyl_zQw2zE . License : All Rights Reserved . License Terms : Standard YouTube License
• Cofunction Identities. Authored by : Mathispower4u. Located at : https://youtu.be/_gkuml--4_Q . License : All Rights Reserved . License Terms : Standard YouTube License
• Example: Determine the Length of a Side of a Right Triangle Using a Trig Equation. Authored by : Mathispower4u. Located at : https://youtu.be/8jU2R3BuR5E . License : All Rights Reserved . License Terms : Standard YouTube License

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Unit 5: Right triangles & trigonometry

About this unit.

Triangles are not always right (although they are never wrong), but when they are it opens up an exciting world of possibilities. Not only are right triangles cool in their own right (pun intended), they are the basis of very important ideas in analytic geometry (the distance between two points in space) and trigonometry.

Pythagorean theorem

• Getting ready for right triangles and trigonometry (Opens a modal)
• Pythagorean theorem in 3D (Opens a modal)
• Pythagorean theorem with isosceles triangle (Opens a modal)
• Multi-step word problem with Pythagorean theorem (Opens a modal)
• Pythagorean theorem in 3D Get 3 of 4 questions to level up!
• Pythagorean theorem challenge Get 3 of 4 questions to level up!

Pythagorean theorem proofs

• Garfield's proof of the Pythagorean theorem (Opens a modal)
• Bhaskara's proof of the Pythagorean theorem (Opens a modal)
• Pythagorean theorem proof using similarity (Opens a modal)
• Another Pythagorean theorem proof (Opens a modal)

Special right triangles

• Special right triangles proof (part 1) (Opens a modal)
• Special right triangles proof (part 2) (Opens a modal)
• 30-60-90 triangle example problem (Opens a modal)
• Area of a regular hexagon (Opens a modal)
• Special right triangles review (Opens a modal)
• Special right triangles Get 3 of 4 questions to level up!

Ratios in right triangles

• Hypotenuse, opposite, and adjacent (Opens a modal)
• Side ratios in right triangles as a function of the angles (Opens a modal)
• Using similarity to estimate ratio between side lengths (Opens a modal)
• Using right triangle ratios to approximate angle measure (Opens a modal)
• Use ratios in right triangles Get 3 of 4 questions to level up!

Introduction to the trigonometric ratios

• Triangle similarity & the trigonometric ratios (Opens a modal)
• Trigonometric ratios in right triangles (Opens a modal)
• Trigonometric ratios in right triangles Get 3 of 4 questions to level up!

Solving for a side in a right triangle using the trigonometric ratios

• Solving for a side in right triangles with trigonometry (Opens a modal)
• Solve for a side in right triangles Get 3 of 4 questions to level up!

Solving for an angle in a right triangle using the trigonometric ratios

• Intro to inverse trig functions (Opens a modal)
• Solve for an angle in right triangles Get 3 of 4 questions to level up!

Sine & cosine of complementary angles

• Sine & cosine of complementary angles (Opens a modal)
• Using complementary angles (Opens a modal)
• Trig word problem: complementary angles (Opens a modal)
• Trig challenge problem: trig values & side ratios (Opens a modal)
• Relate ratios in right triangles Get 3 of 4 questions to level up!

Modeling with right triangles

• Right triangle word problem (Opens a modal)
• Angles of elevation and depression (Opens a modal)
• Right triangle trigonometry review (Opens a modal)
• Right triangles and trigonometry FAQ (Opens a modal)
• Right triangle trigonometry word problems Get 3 of 4 questions to level up!

Trigonometry Worksheets

Related Pages Math Worksheets Free Printable Worksheets

There are six sets of Trigonometry worksheets:

• Trig Ratios: Sin, Cos, Tan
• Sin & Cos of Complementary Angles
• Find Missing Sides
• Find Missing Angles
• Area of Triangle using Sine
• Law of Sines and Cosines

Sine, Cosine, & Tangent Worksheets

In these free math worksheets, students learn how to find the trig ratios: sine, cosine, and tangent.

There are five sets of Sine, Cosine, & Tangent worksheets: Trigonometry Worksheet (Learn Adjacent, Opposite, & Hypotenuse) Trigonometric Ratios Worksheets (Sine Ratio, Cosine Ratio, Tangent Ratio)

How to find Sine, Cosine, & Tangent worksheets? In the context of trigonometry, the sides of a right triangle are often described in relation to an angle within the triangle. The common terms used are:

Hypotenuse: The side opposite the right angle. It is the longest side of the right triangle.

Opposite: The side opposite a specified angle. In other words, if you’re looking at one of the non-right angles, the side opposite that angle is called the “opposite” side.

Adjacent: The side adjacent to a specified angle. It is the side that is next to the angle but is not the hypotenuse.

Sine (sin), cosine (cos), and tangent (tan) are three fundamental trigonometric functions that describe the relationships between the sides and angles of a right triangle. These functions are widely used in mathematics and various scientific fields.

Sine (sin): In a right triangle, the sine of an angle (θ) is the ratio of the length of the side opposite the angle to the length of the hypotenuse. sin(θ)= Opposite/Hypotenuse

Cosine (cos): The cosine of an angle (θ) in a right triangle is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse. cos(θ)= Adjacent/Hypotenuse

Tangent (tan): The tangent of an angle (θ) in a right triangle is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. tan(θ)= Opposite/Adjacent

These functions are defined not only in the context of right triangles but also more broadly using the unit circle and as periodic functions. They have various applications in physics, engineering, computer science, and other fields where understanding the relationships between angles and sides is crucial.

Click on the following worksheet to get a printable pdf document. Scroll down the page for more Sine, Cosine, & Tangent Worksheets .

More Sine, Cosine, & Tangent Worksheets

Printable (Answers on the second page.) Trigonometry Worksheet #1 (Adjacent, Opposite, & Hypotenuse) Trig Ratios Worksheet #2 (Sin, Cos, Tan Ratios) Trig Ratios Worksheet #3 (Sin, Cos, Tan Ratios) Trig Ratios Worksheet #4 (Sin, Cos, Tan Ratios) Trig Ratios Worksheet #5 (Sin, Cos, Tan Ratios)

Online Trigonometry (sine, cosine, tangent) Trigonometry (sine, cosine, tangent) Trigonometry (using a calculator) Inverse Trigonometry (using a calculator) Trigonometry (find an unknown side) Trigonometry (find an unknown angle) Using Sine Using Cosine Using Tangent Using Sine, Cosine or Tangent Trigonometry Applications Problems Law of Sines or Sine Rule Law of Sines Law of Cosines or Cosine Rule Law of Cosines

Related Lessons & Worksheets

Trigonometry Lessons

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