Trigonometry Textbook Exercise - Corbettmaths - Free Printable

To solve the problem of finding the missing angles in the given right triangles, we will use trigonometric ratios. The three primary trigonometric ratios are:

1. Sine (sin): $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
2. Cosine (cos): $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
3. Tangent (tan): $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$

We will determine which ratio to use based on the sides provided in each triangle.

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Question 1: Find the size of the missing angles in the triangles below.

#### (a)
- Given:
- Opposite side = 3 cm
- Adjacent side = 6 cm
- Hypotenuse is not directly given, but we can use the tangent ratio since we have the opposite and adjacent sides.

- Step 1: Use the tangent ratio:
$$
\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{6} = 0.5
$$

- Step 2: Solve for \( x \):
$$
x = \tan^{-1}(0.5)
$$
Using a calculator:
$$
x \approx 26.57^\circ
$$

#### (b)
- Given:
- Opposite side = 4 cm
- Hypotenuse = 15 cm
- Adjacent side is not directly given, but we can use the sine ratio since we have the opposite and hypotenuse.

- Step 1: Use the sine ratio:
$$
\sin x = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{15}
$$

- Step 2: Solve for \( x \):
$$
x = \sin^{-1}\left(\frac{4}{15}\right)
$$
Using a calculator:
$$
x \approx 15.34^\circ
$$

#### (c)
- Given:
- Opposite side = 8 cm
- Hypotenuse = 10 cm
- Adjacent side is not directly given, but we can use the sine ratio since we have the opposite and hypotenuse.

- Step 1: Use the sine ratio:
$$
\sin x = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8}{10} = 0.8
$$

- Step 2: Solve for \( x \):
$$
x = \sin^{-1}(0.8)
$$
Using a calculator:
$$
x \approx 53.13^\circ
$$

#### (d)
- Given:
- Opposite side = 9 cm
- Adjacent side = 12 cm
- Hypotenuse is not directly given, but we can use the tangent ratio since we have the opposite and adjacent sides.

- Step 1: Use the tangent ratio:
$$
\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{9}{12} = 0.75
$$

- Step 2: Solve for \( x \):
$$
x = \tan^{-1}(0.75)
$$
Using a calculator:
$$
x \approx 36.87^\circ
$$

#### (e)
- Given:
- Opposite side = 7 cm
- Adjacent side = 20 cm
- Hypotenuse is not directly given, but we can use the tangent ratio since we have the opposite and adjacent sides.

- Step 1: Use the tangent ratio:
$$
\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{20} = 0.35
$$

- Step 2: Solve for \( x \):
$$
x = \tan^{-1}(0.35)
$$
Using a calculator:
$$
x \approx 19.29^\circ
$$

#### (f)
- Given:
- Opposite side = 42 cm
- Adjacent side = 48 cm
- Hypotenuse is not directly given, but we can use the tangent ratio since we have the opposite and adjacent sides.

- Step 1: Use the tangent ratio:
$$
\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{42}{48} = 0.875
$$

- Step 2: Solve for \( x \):
$$
x = \tan^{-1}(0.875)
$$
Using a calculator:
$$
x \approx 41.19^\circ
$$

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Final Answers:

$$
\boxed{
\begin{aligned}
(a) & \quad x \approx 26.57^\circ \\
(b) & \quad x \approx 15.34^\circ \\
(c) & \quad x \approx 53.13^\circ \\
(d) & \quad x \approx 36.87^\circ \\
(e) & \quad x \approx 19.29^\circ \\
(f) & \quad x \approx 41.19^\circ \\
\end{aligned}
}
$$