Trigonometry worksheet for finding missing sides and angles in right triangles.

Here are the solutions for the missing sides in the right triangles. I have rounded decimal answers to two places (nearest hundredth) as is standard for these problems.

1)
* Find $x$ (hypotenuse): We know the side adjacent to $70^\circ$ is 7.2. We use Cosine.
$\cos(70^\circ) = \frac{7.2}{x}$
$x = \frac{7.2}{\cos(70^\circ)} \approx 21.05$
* Find $y$ (opposite): We use Tangent.
$\tan(70^\circ) = \frac{y}{7.2}$
$y = 7.2 \cdot \tan(70^\circ) \approx 19.78$

2)
* Find $x$ (adjacent): We know the hypotenuse is 20 and the angle is $18^\circ$. We use Cosine.
$\cos(18^\circ) = \frac{x}{20}$
$x = 20 \cdot \cos(18^\circ) \approx 19.02$
* Find $y$ (opposite): We use Sine.
$\sin(18^\circ) = \frac{y}{20}$
$y = 20 \cdot \sin(18^\circ) \approx 6.18$

3)
* Find $x$ (opposite): We know the adjacent side is 5 and the angle is $50^\circ$. We use Tangent.
$\tan(50^\circ) = \frac{x}{5}$
$x = 5 \cdot \tan(50^\circ) \approx 5.96$
* Find $y$ (hypotenuse): We use Cosine.
$\cos(50^\circ) = \frac{5}{y}$
$y = \frac{5}{\cos(50^\circ)} \approx 7.78$

4)
* Find $x$ (opposite): We know the adjacent side is 10 and the angle is $64^\circ$. We use Tangent.
$\tan(64^\circ) = \frac{x}{10}$
$x = 10 \cdot \tan(64^\circ) \approx 20.50$
* Find $y$ (hypotenuse): We use Cosine.
$\cos(64^\circ) = \frac{10}{y}$
$y = \frac{10}{\cos(64^\circ)} \approx 22.81$

5)
* This is a special $45^\circ-45^\circ-90^\circ$ triangle, so the legs are equal ($x = y$).
* Find $x$: We know the hypotenuse is 4.
$\sin(45^\circ) = \frac{x}{4}$
$x = 4 \cdot \sin(45^\circ) \approx 2.83$
* Find $y$: Since it's isosceles, $y = x$.
$y \approx 2.83$

6)
* Find $x$ (opposite): We know the hypotenuse is 1 and the angle is $30^\circ$. We use Sine.
$\sin(30^\circ) = \frac{x}{1}$
$x = 1 \cdot 0.5 = 0.5$
* Find $y$ (adjacent): We use Cosine.
$\cos(30^\circ) = \frac{y}{1}$
$y = 1 \cdot \cos(30^\circ) \approx 0.87$

7)
* Find $x$ (adjacent): We know the opposite side is 27 and the angle is $30^\circ$. We use Tangent.
$\tan(30^\circ) = \frac{27}{x}$
$x = \frac{27}{\tan(30^\circ)} \approx 46.77$
* Find $y$ (hypotenuse): We use Sine.
$\sin(30^\circ) = \frac{27}{y}$
$y = \frac{27}{\sin(30^\circ)} = \frac{27}{0.5} = 54$

8)
* Find $x$ (hypotenuse): We know the adjacent side to the $60^\circ$ angle is 6. We use Cosine.
$\cos(60^\circ) = \frac{6}{x}$
$x = \frac{6}{\cos(60^\circ)} = \frac{6}{0.5} = 12$
* Find $y$ (opposite): We use Tangent.
$\tan(60^\circ) = \frac{y}{6}$
$y = 6 \cdot \tan(60^\circ) \approx 10.39$

Final Answer:
1) x ≈ 21.05, y ≈ 19.78
2) x ≈ 19.02, y ≈ 6.18
3) x ≈ 5.96, y ≈ 7.78
4) x ≈ 20.50, y ≈ 22.81
5) x ≈ 2.83, y ≈ 2.83
6) x = 0.5, y ≈ 0.87
7) x ≈ 46.77, y = 54
8) x = 12, y ≈ 10.39