We are given two right triangles and asked to find the value of the angle $x$ in each triangle using trigonometric ratios. For the first triangle, the side opposite angle $x$ is 11, and the hypotenuse is 20. For the second triangle, the side opposite angle $x$ is 16, and the adjacent side is 7.

1. Problem Description

We are given two right triangles and asked to find the value of the angle $x$ in each triangle using trigonometric ratios. For the first triangle, the side opposite angle $x$ is 11, and the hypotenuse is

2

0. For the second triangle, the side opposite angle $x$ is 16, and the adjacent side is

7.

2. Solution Steps

For the first triangle:

We have the opposite and hypotenuse, so we use the sine function:

$sin(x) = \frac{opposite}{hypotenuse}$

$sin(x) = \frac{11}{20}$

$x = arcsin(\frac{11}{20})$

$x \approx 33.367^\circ$

To the nearest degree, $x = 33^\circ$.

For the second triangle:

We have the opposite and adjacent, so we use the tangent function:

$tan(x) = \frac{opposite}{adjacent}$

$tan(x) = \frac{16}{7}$

$x = arctan(\frac{16}{7})$

$x \approx 66.356^\circ$

To the nearest degree, $x = 66^\circ$.

3. Final Answer

For the first triangle: $x = 33^\circ$

For the second triangle: $x = 66^\circ$