Find the value of the unknown sides in each of the following right angled triangles. Use trigonometric ratios to find the size of the angle marked theta and beta in each of the following right-angled triangles.

Here are the solutions to the problems using trigonometric ratios.

1. Find the value of the unknown sides in each of the following right-angled triangles.

a) For triangle ABC: Step 1: To find $y$, use the sine ratio. $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ $\sin(10^\circ) = \frac{y}{16 \text{ cm}}$ $y = 16 \sin(10^\circ)$ $y \approx 16 \times 0.1736$ $y \approx 2.78 \text{ cm}$ Step 2: To find $x$, use the cosine ratio. $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$ $\cos(10^\circ) = \frac{x}{16 \text{ cm}}$ $x = 16 \cos(10^\circ)$ $x \approx 16 \times 0.9848$ $x \approx 15.76 \text{ cm}$ The values are: $\boxed{y \approx 2.78 \text{ cm}, x \approx 15.76 \text{ cm}}$

b) For triangle XYZ: Step 1: To find $a$, use the sine ratio. $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ $\sin(35^\circ) = \frac{5 \text{ cm}}{a}$ $a = \frac{5 \text{ cm}}{\sin(35^\circ)}$ $a \approx \frac{5}{0.5736}$ $a \approx 8.72 \text{ cm}$ Step 2: To find $b$, use the tangent ratio. $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ $\tan(35^\circ) = \frac{5 \text{ cm}}{b}$ $b = \frac{5 \text{ cm}}{\tan(35^\circ)}$ $b \approx \frac{5}{0.7002}$ $b \approx 7.14 \text{ cm}$ The values are: $\boxed{a \approx 8.72 \text{ cm}, b \approx 7.14 \text{ cm}}$

c) For triangle PQR: Step 1: To find $y$, use the sine ratio. $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ $\sin(60^\circ) = \frac{6 \text{ cm}}{y}$ $y = \frac{6 \text{ cm}}{\sin(60^\circ)}$ $y \approx \frac{6}{0.8660}$ $y \approx 6.93 \text{ cm}$ Step 2: To find $x$, use the tangent ratio. $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ $\tan(60^\circ) = \frac{6 \text{ cm}}{x}$ $x = \frac{6 \text{ cm}}{\tan(60^\circ)}$ $x \approx \frac{6}{1.7321}$ $x \approx 3.46 \text{ cm}$ The values are: $\boxed{x \approx 3.46 \text{ cm}, y \approx 6.93 \text{ cm}}$

d) For triangle KLM: Step 1: To find $x$, use the sine ratio. $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ $\sin(68.7^\circ) = \frac{x}{25 \text{ cm}}$ $x = 25 \sin(68.7^\circ)$ $x \approx 25 \times 0.9318$ $x \approx 23.30 \text{ cm}$ Step 2: To find $y$, use the cosine ratio. $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$ $\cos(68.7^\circ) = \frac{y}{25 \text{ cm}}$ $y = 25 \cos(68.7^\circ)$ $y \approx 25 \times 0.3633$ $y \approx 9.08 \text{ cm}$ The values are: $\boxed{x \approx 23.30 \text{ cm}, y \approx 9.08 \text{ cm}}$

2. Use trigonometric ratios to find the size of the angle marked $\theta$ and $\beta$ in each of the following right-angled triangles.

a) For the first triangle: Step 1: To find $\theta$, use the tangent ratio. $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ $\tan(\theta) = \frac{5 \text{ cm}}{8 \text{ cm}}$ $\theta = \arctan\left(\frac{5}{8}\right)$ $\theta \approx 32.0^\circ$ Step 2: To find $\beta$, use the fact that the sum of acute angles in a right triangle is $90^\circ$. $\beta = 90^\circ - \theta$ $\beta = 90^\circ - 32.0^\circ$ $\beta \approx 58.0^\circ$ The angles are: $\boxed{\theta \approx 32.0^\circ, \beta \approx 58.0^\circ}$

b) For the second triangle: Step 1: To find $\theta$, use the sine ratio. $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ $\sin(\theta) = \frac{4 \text{ cm}}{10 \text{ cm}}$ $\sin(\theta) = 0.4$ $\theta = \arcsin(0.4)$ $\theta \approx 23.6^\circ$ Step 2: To find $\beta$, use the fact that the sum of acute angles in a right triangle is $90^\circ$. $\beta = 90^\circ - \theta$ $\beta = 90^\circ - 23.6^\circ$ $\beta \approx 66.4^\circ$ The angles are: $\boxed{\theta \approx 23.6^\circ, \beta \approx 66.4^\circ}$

c) For the third triangle: Step 1: To find $\theta$, use the sine ratio. $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ $\sin(\theta) = \frac{16 \text{ cm}}{25 \text{ cm}}$ $\sin(\theta) = 0.64$ $\theta = \arcsin(0.64)$ $\theta \approx 39.8^\circ$ Step 2: To find $\beta$, use the fact that the sum of acute angles in a right triangle is $90^\circ$. $\beta = 90^\circ - \theta$ $\beta = 90^\circ - 39.8^\circ$ $\beta \approx 50.2^\circ$ The angles are: $\boxed{\theta \approx 39.8^\circ, \beta \approx 50.2^\circ}$