Trigonometry Exercises

Exercise 1

Q1. In a right-angled triangle, the base is 4 cm and the perpendicular is 3 cm.
Find the hypotenuse.

Reveal solution

Solution:

$h = \sqrt{p^2 + b^2}$

= $\sqrt{3^2 + 4^2}$

= $\sqrt{9 + 16}$

= $\sqrt{25}$

= $\text{5cm}$

Exercise 2

Q2. In a right-angled triangle, the perpendicular is 5 cm and the hypotenuse is 13 cm.
Find $( \tan \theta )$.

Reveal solution

Solution:

First, find the base using the Pythagorean theorem:

$h^2 = p^2 + b^2$

$13^2 = 5^2 + b^2$

$169 = 25 + b^2$

$b^2 = 169 - 25 = 144$

$b = 12$ cm

Now,

$\tan \theta = \dfrac{p}{b} = \dfrac{5}{12}$

Exercise 3

Q2. In a triangle, AB = 12cm, BC = 5cm and AC = 13cm. $\angle ACB = \theta$. Find the value of $\theta$.

Reveal solution

Figure 3: Intersection of two sets A and B.

Given:

AB = 12 cm, BC = 5 cm, AC = 13 cm, and $\triangle ABC$ is right-angled at B.

To find: $\angle ACB = \theta$

Since AC is the hypotenuse, and BC is adjacent to angle $\theta$ at C:

$\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{5}{13}$

$\theta = \cos^{-1} \left( \dfrac{5}{13} \right )$

So, $\boxed{\theta = \cos^{-1} \left( \dfrac{5}{13} \right )}$

Or, using a calculator:

$\theta \approx 67.38^\circ$