5.5: Right Triangle Trigonometry

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In Section 5.3 we were introduced to the sine and cosine function as ratios of the sides of a triangle drawn inside a circle, and spent the rest of that section discussing the role of those functions in finding points on the circle. In this section, we return to the triangle, and explore the applications of the trigonometric functions to right triangles where circles may not be involved.

Recall that we defined sine and cosine as $A circle centered at the origin with a line labeled r drawn at an angle of theta. The point where the line meets the circle is labeled x comma y. A vertical line is drawn from that point to the x axis forming a triangle, with the vertical length labeled y. The horizontal leg of the triangle from the origin to the vertical line is labeled x.$

\[\sin (\theta ) = \dfrac{y}{r}\nonumber\]

\[\cos(\theta ) = \dfrac{x}{r}\nonumber\]

Separating the triangle from the circle, we can make equivalent but more general definitions of the sine, cosine, and tangent on a right triangle. On the right triangle, we will label the hypotenuse as well as the side opposite the angle and the side adjacent (next to) the angle.

RIGHT TRIANGLE RELATIONSHIPS

Given a right triangle with an angle of $\theta$

$A right triangle with one angle labeled theta. The leg touching the angle theta is labeled adjacent. The leg across from the angle is labeled opposite. The hypotenuse is labeled hypotenuse.$

\[\text{sin} (\theta) = \dfrac{\text{opposite}}{\text{hypotenuse}}\]

\[\text{cos} (\theta) = \dfrac{\text{adjacent}}{\text{hypotenuse}}\]

\[\text{tan} (\theta) = \dfrac{\text{opposite}}{\text{adjacent}}\]

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “Sine is opposite over hypotenuse, Cosine is adjacent over hypotenuse, Tangent is opposite over adjacent.”

Example $\PageIndex{1}$

$\alpha$$\alpha$Given the triangle shown, find the value for $\text{cos}(\alpha)$. $A right triangle with hypotenuse 17 and legs 8 and 15. The angle between the sides length 15 and 17 is labeled alpha.$

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

\[\text{cos}(\alpha) = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{15}{17}\nonumber\]

When working with general right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the sine and cosine of either of the two acute angles in the triangle.

$A rotated right triangle with angles labeled alpha and beta. The leg touching the corner with angle alpha is labeled adjacent to alpha opposite beta. The leg touching the corner with angle beta is labeled adjacent to beta opposite alpha.$

Example $\PageIndex{2}$

Using the triangle shown, evaluate cos($\alpha$), sin($\alpha$), cos($\beta$), and sin($\beta$),

$A rotated right triangle with sides labeled 3, 4, and 5. The angle between sides length 3 and 5 is labeled alpha. The angle between sides length 4 and 5 is labeled beta.$

Solution

\[\text{cos}(\alpha) = \dfrac{\text{adjacent to } \alpha}{\text{hypotenuse}} = \dfrac{3}{5}\nonumber\]

\[\text{sin}(\alpha) = \dfrac{\text{opposite } \alpha}{\text{hypotenuse}} = \dfrac{4}{5}\nonumber\]

\[\text{cos}(\beta) = \dfrac{\text{adjacent to } \beta}{\text{hypotenuse}} = \dfrac{4}{5}\nonumber\]

\[\text{cos}(\beta) = \dfrac{\text{opposite } \beta}{\text{hypotenuse}} = \dfrac{3}{5}\nonumber\]

Exercise $\PageIndex{1}$

A right triangle is drawn with angle $\alpha$ opposite a side with length 33, angle $\beta$ opposite a side with length 56, and hypotenuse 65. Find the sine and cosine of $\alpha$ and $\beta$

Answer

\[\text{sin} (\alpha) = \frac{33}{65}\nonumber\]

\[\text{cos} (\alpha) = \frac{56}{65}\nonumber\]

\[\text{sin} (\beta) = \frac{56}{65}\nonumber\]

\[\text{cos} (\beta) = \frac{33}{65}\nonumber\]

You may have noticed that in the above example that $\cos (\alpha )=\sin (\beta )$ and $\cos (\beta )=\sin (\alpha )$. This makes sense since the side opposite $\alpha$ is also adjacent to $\beta$. Since the three angles in a triangle need to add to $\pi$, or 180 degrees, then the other two angles must add to $\dfrac{\pi }{2}$, or 90 degrees, so $\beta =\dfrac{\pi }{2} -\alpha$, and $\alpha =\dfrac{\pi }{2} -\beta$. Since $\cos (\alpha )=\sin (\beta )$, then $\cos (\alpha )=\sin \left(\dfrac{\pi }{2} -\alpha \right)$.

COFUNCTION IDENTITIES

The confunction identities for sine and cosine are:

\[\text{cos} (\theta) = \text{sin} (\dfrac{\pi}{2} - \theta)\]

\[\text{sin} (\theta) = \text{cos} (\dfrac{\pi}{2} - \theta)\]

In the previous examples, we evaluated the sine and cosine on triangles where we knew all three sides of the triangle. Right triangle trigonometry becomes powerful when we start looking at triangles in which we know an angle but don’t know all the sides.

Example $\PageIndex{3}$

Find the unknown sides of the triangle pictured here.

$A right triangle. The hypotenuse is labeled b and the legs are labeled a and 7. The angle between the sides length a and b is 30 degrees.$

Solution

Since $\text{sin} (\theta) = \dfrac{\text{opposite}}{\text{hypotenuse}}$. $\text{sin}(30^{\circ}) = \dfrac{7}{b}$.

From this, we can solve for the side $b$.

\[b\sin (30{}^\circ )=7\nonumber\]
\[b=\dfrac{7}{\sin (30{}^\circ )}\nonumber\]

To obtain a value, we can evaluate the sine and simplify

\[b = \dfrac{7}{1/2} = 14\nonumber\]

To find the value for side $a$, we could use the cosine, or simply apply the Pythagorean Theorem:

\[a^2 + 7^2 = b^2\nonumber\]
\[a^2 + 7^2 = 14^2\nonumber\]
\[a = \sqrt{147}\nonumber\]

Notice that if we know at least one of the non-right angles of a right triangle and one side, we can find the rest of the sides and angles.

Exercise $\PageIndex{2}$

A right triangle has one angle of $\dfrac{\pi }{3}$ and a hypotenuse of 20. Find the unknown sides and angles of the triangle.

Answer

\[\text{cos} (\dfrac{\pi}{3}) = \dfrac{\text{adjacent}}{\text{hypoteuse}} = \dfrac{\text{Adj}}{20}\nonumber\] so, \[\text{adjacent} = 20 \text{cos} (\dfrac{\pi}{3}) = 20 (\dfrac{1}{2}) = 10\nonumber\]

\[\text{sin} (\dfrac{\pi}{3}) = \dfrac{\text{Opposite}}{\text{hypotenuse}} = \dfrac{\text{Opp}}{20}\nonumber\] so, \[\text{opposite} = 20 \text{sin} (\dfrac{\pi}{3}) = 20 (\dfrac{\sqrt{3}}{2}) = 10\sqrt{3}\nonumber\]

Missing angle = 180 - 90 - 60 = 30 degrees or $\pi / 6$.

Example $\PageIndex{4}$

To find the height of a tree, a person walks to a point 30 feet from the base of the tree, and measures the angle from the ground to the top of the tree to be 57 degrees. Find the height of the tree.

Solution

We can introduce a variable, $h$, to represent the height of the tree. The $A picture of a tree. A point is shown a horizontal distance of 30 feet to the left of the tree. From that point a line is drawn to the top of the tree, with the angle between the line and horizontal is 57 degrees.$ two sides of the triangle that are most important to us are the side opposite the angle, the height of the tree we are looking for, and the adjacent side, the side we are told is 30 feet long.

The trigonometric function which relates the side opposite of the angle and the side adjacent to the angle is the tangent.

\[\text{tan} (57^{\circ}) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{h}{30}\nonumber\] Solving for $h$,

\[h=30\tan (57{}^\circ )\nonumber\] Using technology, we can approximate a value

\[h=30\tan (57{}^\circ )\approx 46.2\text{ feet}\nonumber\]

The tree is approximately 46 feet tall.

Example $\PageIndex{5}$

A person standing on the roof of a 100 foot tall building is looking towards a skyscraper a few blocks away, wondering how tall it is. She measures the angle of declination from the roof of the building to the base of the skyscraper to be 20 degrees and the angle of inclination to the top of the skyscraper to be 42 degrees.

Solution

To approach this problem, it would be good to start with a picture. Although we are interested in the height, $Two building are shown. The building on the left is 100 feet tall; the building on the right has height labeled h. From the top of the first building a horizontal line is drawn right to the other building with length labeled x. A line is drawn from the top of the first building to the bottom of the second building, at an angle of 20 degrees below horizontal. A line is drawn from the top of the first building to the top of the second building, at an angle of 42 degrees. The vertical distance from the horizontal line to the top of the second building is labeled a.$ $h$, of the skyscraper, it can be helpful to also label other unknown quantities in the picture – in this case the horizontal distance $x$ between the buildings and $a$, the height of the skyscraper above the person.

To start solving this problem, notice we have two right triangles. In the top triangle, we know one angle is 42 degrees, but we don’t know any of the sides of the triangle, so we don’t yet know enough to work with this triangle.

In the lower right triangle, we know one angle is 20 degrees, and we know the vertical height measurement of 100 ft. Since we know these two pieces of information, we can solve for the unknown distance $x$.

\[\text{tan} (20^{\circ}) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{100}{x}\nonumber\] Solving for $x$
\[x \text{tan} (20^{\circ}) = 100\nonumber\]
\[x = \dfrac{100}{\text{tan} (20^{\circ})}\nonumber\]

Now that we have found the distance $x$, we know enough information to solve the top right triangle.

\[\text{tan} (42^{\circ}) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{a}{x} = \dfrac{a}{100/\text{tan}(20^{circ})}\nonumber\]
\[\text{tan} (42^{\circ}) = \dfrac{a\text{tan} (20^{\circ})}{100}\nonumber\]
\[100 \text{tan} (42^{\circ}) = a\text{tan} (20^{\circ})\nonumber\]
\[\dfrac{\text{tan} (42^{\circ})}{\text{tan} (20^{\circ})} = a\nonumber\]

Approximating a value,

\[a=\dfrac{100\tan (42{}^\circ )}{\tan (20{}^\circ )} \approx 247.4\text{ feet}\nonumber\]

Adding the height of the first building, we determine that the skyscraper is about 347 feet tall.

Important Topics of This Section

SOH CAH TOA
Cofunction identities
Applications with right triangles

Example $\PageIndex{3}$

Solve the triangle for the angle $\theta$.

$A right triangle with hypotenuse labeled 12 and one side labeled 9, and the angle between them labeled theta.$

Solution

Since we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.

\[\cos \left(\theta \right)=\dfrac{9}{12}\nonumber\]Using the definition of the inverse,

\[\theta =\cos ^{-1} \left(\dfrac{9}{12} \right)\nonumber\]Evaluating

\[\theta \approx 0.7227\text{, or about }41.4096\mathrm{{}^\circ}\nonumber\]

There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can find exact values for the resulting expressions

Example $\PageIndex{6}$

Find an exact value for $\sin \left(\tan ^{-1} \left(\dfrac{7}{4} \right)\right)$.

Solution

While we could use a similar te $A right triangle with legs 4 and 7. The side length 4 meets the hypotenuse at an angle labeled theta.$ chnique as in the last example, we will demonstrate a different technique here. From the inside, we know there is an angle so $\tan \left(\theta \right)=\dfrac{7}{4}$. We can envision this as the opposite and adjacent sides on a right triangle.

Using the Pythagorean Theorem, we can find the hypotenuse of this triangle:

\[4^{2} +7^{2} =hypotenuse ^{2}\nonumber\]

\[hypotenuse=\sqrt{65}\nonumber\]

Now, we can represent the sine of the angle as opposite side divided by hypotenuse.

\[\sin \left(\theta \right)=\dfrac{7}{\sqrt{65} }\nonumber\]

This gives us our desired composition

\[\sin \left(\tan ^{-1} \left(\dfrac{7}{4} \right)\right)=\sin (\theta )=\dfrac{7}{\sqrt{65} } .\nonumber\]

Solving right triangles for angles

In Section 5.5, we used trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trig functions, we can solve for the angles of a right triangle given two sides.

Example $\PageIndex{1}$

An airplane needs to fly to an airfield located 300 miles east and 200 miles north of its current location. At what heading should the airplane fly? In other words, if we ignore air resistance or wind speed, how many degrees north of east should the airplane fly?

Solution

We might begin by drawing a picture and labeling all of the known information.

$A right triangle with legs length 200 and 300. The angle between the hypotenuse and leg length 300 is labeled alpha.$

Drawing a triangle, we see we are looking for the angle $\alpha$. In this triangle, the side opposite the angle $\alpha$ is 200 miles and the side adjacent is 300 miles. Since we know the values for the opposite and adjacent sides, it makes sense to use the tangent function.

\[\tan (\alpha )=\dfrac{200}{300} \nonumber\]

Using the inverse,

\[\alpha =\tan ^{-1} \left(\dfrac{200}{300} \right)\approx 0.588 \nonumber\]

or equivalently about 33.7 degrees.

The airplane needs to fly at a heading of 33.7 degrees north of east.

Example $\PageIndex{2}$

OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length (http://www.osha.gov/SLTC/etools/cons.../4ladders.html). Find the angle such a ladder forms with the ground.

Solution

For any length of ladder, the base needs to be one quarter of the distance the foot of the ladder is away from the wall. Equivalently, if the base is $a$ feet from the wall, the ladder can be $4a$ feet long. Since $a$ is the side adjacent to the angle and $4a$ is the hypotenuse, we use the cosine function.a $A right triangle with leg length a and hypotenuse length 4a and angle between labeled theta.$

\[\cos (\theta )=\dfrac{a}{4a} =\dfrac{1}{4} \nonumber\]

Using the inverse

\[\theta =\cos ^{-1} \left(\dfrac{1}{4} \right)\approx 75.52\, \text{degrees}\nonumber\]

The ladder forms a 75.52 degree angle with the ground.

Exercise $\PageIndex{1}$

A cable that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 70 meters above the ground and the second anchor on the ground is 23 meters from the base of the wheel. What is the angle from the ground up to the center of the Ferris wheel and how long is the cable?

Answer: Angle of elevation for the cable is 71.81 degrees and the cable is 73.68 m long

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