Precalculus by Richard Wright

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4-03 Right Triangle Trigonometry

A 8-foot step ladder actually is not 8-feet high. The size of a step ladder is actually the length of the rails the steps are attached to. When the ladder is in use, the rails are slanted so the height is less. Using right triangles we can calculate the actual height of an 8-foot step ladder.

Right Triangle Trigonometry

A right triangle has one right angle and two acute angles. The side opposite the right angle is the hypotenuse and is the longest side. The other sides are called legs. If one of the acute angles is chosen the leg forming one side of the angle is called the adjacent leg. The leg opposite from the chosen angle is called the opposite leg.

The six trigonometric functions have definitions similar to the unit circle, only this time the angle must be acute and the fraction part of the formulas is a ratio of sides of the triangle.

Notice that sine and cosecant, cosine and secant, tangent and cotangent are reciprocals just like they were with the unit circle.

Evaluate Trigonometric Functions

Evaluate the six trigonometric functions of α for the triangle in figure 3.

Solution

Use the Pythagorean Theorem to solve for the third side.

$\begin{array}{l}{\left(\text{adj}\right)}^2+{\left(\text{opp}\right)}^2={\left(\text{hyp}\right)}^2\\ 8^2+{\left(\text{opp}\right)}^2={10}^2\\ opp=\sqrt{{10}^2-8^2}\\ opp=\sqrt{36}\\ opp=6\end{array}$

From the figure, adj = 8, hyp = 10, and opp = 6. Use the formulas to evaluate the trigonometric functions.

$sin\alpha =\frac{opp}{hyp}=\frac{6}{10}=\frac{3}{5}$	$csc\alpha =\frac{opp}{hyp}=\frac{10}{6}=\frac{5}{3}$
$cos\alpha =\frac{adj}{hyp}=\frac{8}{10}=\frac{4}{5}$	$sec\alpha =\frac{opp}{hyp}=\frac{10}{8}=\frac{5}{4}$
$tan\alpha =\frac{opp}{adj}=\frac{6}{8}=\frac{3}{4}$	$cot\alpha =\frac{adj}{opp}=\frac{8}{6}=\frac{4}{3}$

Evaluate the six trigonometric functions of β for the triangle in figure 4.

Answers

$sin\beta =\frac{2\sqrt{13}}{13}$	$csc\beta =\frac{\sqrt{13}}{2}$
$cos\beta =\frac{3\sqrt{13}}{13}$	$sec\beta =\frac{\sqrt{13}}{3}$
$tan\beta =\frac{2}{3}$	$cot\alpha =\frac{3}{2}$

Special Right Triangles

Two special triangles contain the most common angles of 30°, 45°, and 60°. The special triangles can be used to evaluate the trigonometric functions of those angles.

45°-45°-90° Triangle

Evaluate the six trigonometric functions of 45°.

Solution

Draw a right triangle with an acute angle of 45°. The acute angles of a right triangle are complementary, so the other acute angle is 45°. Thus this is an isosceles triangle and both legs are the same length. Use the Pythagorean Theorem to find the length of the hypotenuse.

$\begin{array}{l}{hyp}^2=1^2+1^2\\ hyp=\sqrt{2}\end{array}$

$sin45°=\frac{opp}{hyp}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$	$csc45°=\frac{opp}{hyp}=\frac{\sqrt{2}}{1}=\sqrt{2}$
$cos45°=\frac{adj}{hyp}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$	$sec45°=\frac{opp}{hyp}=\frac{\sqrt{2}}{1}=\sqrt{2}$
$tan45°=\frac{opp}{adj}=\frac{1}{1}=1$	$cot45°=\frac{adj}{opp}=\frac{1}{1}=1$

30°-60°-90° Triangle

Evaluate the six trigonometric functions of 30°.

Solution

Draw an equilateral triangle with an angles of 60°. Draw a vertical line to bisect the triangle and create a 30°-60°-90° triangle. Set each side of the equilateral triangle as 2, so the short leg of the 30°-60°-90° triangle is 1. Use the Pythagorean Theorem to find the length of the longer leg.

$\begin{array}{l}{x}^2+1^2=2^2\\ x=\sqrt{2^2-1^2}\\ x=\sqrt{3}\end{array}$

$sin30°=\frac{opp}{hyp}=\frac{1}{2}$	$csc30°=\frac{opp}{hyp}=\frac{2}{1}=2$
$cos30°=\frac{adj}{hyp}=\frac{\sqrt{3}}{2}$	$sec30°=\frac{opp}{hyp}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$
$tan30°=\frac{opp}{adj}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$	$cot30°=\frac{adj}{opp}=\frac{\sqrt{3}}{1}=\sqrt{3}$

Practice Exercises

1. Draw a right triangle and label one acute angle θ. Label the adjacent, opposite, and hypotenuse.
2. Evaluate the six trigonometric functions for the indicated angles.

3. 

4. 

5. 

6. 

7. 

8. 

9. Use the special right triangles to evaluate the indicated trigonometric function.
10. sin 30°
11. csc 45°
12. cot 60°
13. sec 30°
14. $cos\frac{\pi }{4}$
15. $sec\frac{\pi }{3}$
16. $cot\frac{\pi }{4}$
17. $csc\frac{\pi }{6}$
18. Mixed Review
19. (4-02) Using the unit circle, evaluate $sec\frac{3\pi }{2}$.
20. (4-02) Using the unit circle, evaluate sin 570°.
21. (4-01) Draw the angle, $\frac{7\pi }{4}$ in standard position, then find a positive and negative coterminal angle.
22. (3-04) Solve log(x) − log(x + 2) = 1.
23. (2-01) Divide $\frac{2-i}{i}$.

Answers

1. 

2. $sin\alpha =\frac{5}{13},cos\alpha =\frac{12}{13},tan\alpha =\frac{5}{12},csc\alpha =\frac{13}{5},sec\alpha =\frac{13}{12},cot\alpha =\frac{12}{5}$
3. $sin\alpha =\frac{4}{5},cos\alpha =\frac{3}{5},tan\alpha =\frac{4}{3},csc\alpha =\frac{5}{4},sec\alpha =\frac{5}{3},cot\alpha =\frac{3}{4}$
4. $sin\beta =\frac{15}{17},cos\beta =\frac{8}{17},tan\beta =\frac{15}{8},csc\beta =\frac{17}{15},sec\beta =\frac{17}{8},cot\beta =\frac{8}{15}$
5. $sin\beta =\frac{6\sqrt{61}}{61},cos\beta =\frac{5\sqrt{61}}{61},tan\beta =\frac{6}{5},csc\beta =\frac{\sqrt{61}}{6},sec\beta =\frac{\sqrt{61}}{5},cot\beta =\frac{5}{6}$
6. $sin\alpha =\frac{5\sqrt{41}}{41},cos\alpha =\frac{4\sqrt{41}}{41},tan\alpha =\frac{5}{4},csc\alpha =\frac{\sqrt{41}}{5},sec\alpha =\frac{\sqrt{41}}{4},cot\alpha =\frac{4}{5}$
7. $sin\beta =\frac{2\sqrt{22}}{13},cos\beta =\frac{9}{13},tan\beta =\frac{2\sqrt{22}}{9},csc\beta =\frac{13\sqrt{22}}{44},sec\beta =\frac{13}{9},cot\beta =\frac{9\sqrt{22}}{44}$
8. 1⁄2
9. $\sqrt{2}$
10. $\frac{\sqrt{3}}{3}$
11. $\frac{2\sqrt{3}}{3}$
12. $\frac{\sqrt{2}}{2}$
13. 2
14. 1
15. 2
16. undefined
17. $-\frac{1}{2}$

18. 

    ; $\frac{15\pi }{4}$; $-\frac{\pi }{4}$
19. No solution
20. −1 − 2i