# Given sec θ = 13/12, calculate all other trigonometric ratios

**Solution:**

We use the basic concept of trigonometric ratios to solve the problem.

Using sec θ, we can find the ratio of the length of two sides of the right-angled triangle. Then by using the Pythagoras theorem, the third side.

Let ΔABC be a right-angled triangle, right angled at point B.

It is given that:

sec θ = hypotenuse/side adjacent to ∠θ = AC/AB = 13/12

Let AC = 13 k and AB = 12 k, where k is a positive integer.

Apply Pythagoras theorem in Δ ABC, we obtain:

AC^{2} = AB^{2} + BC^{2}

BC^{2} = AC^{2} - AB^{2}

BC^{2} = (13k)^{2} - (12k)^{2}

BC^{2} = 169 k^{2} - 144 k^{2}

BC^{2} = 25k^{2}

BC = 5k

sin θ = side opposite to ∠θ/hypotenuse = BC/AC = 5/13

cos θ = side adjacent to ∠θ/hypotenuse = AB/AC = 12/13

tan θ = side opposite to ∠θ/side adjacent to ∠θ = BC/AB = 5/12

cot θ = side adjacent to ∠θ/side opposite to ∠θ = AB/BC = 12/5

cosec θ = hypotenuse/side opposite to ∠θ = AC/BC = 13/5

**Video Solution:**

## Given sec θ = 13/12, calculate all other trigonometric ratios

### Maths NCERT Solutions Class 10 - Chapter 8 Exercise 8.1 Question 5:

Given sec θ = 13/12, calculate all other trigonometric ratios

If sec θ = 13/12, sin θ = side opposite to ∠θ/hypotenuse = BC/AC = 5/13, cos θ = side adjacent to ∠θ/hypotenuse = AB/AC = 12/13, tan θ = side opposite to ∠θ/side adjacent to ∠θ = BC/AB = 5/12, cot θ = side adjacent to ∠θ/side opposite to ∠θ = AB/BC = 12/5, cosec θ = hypotenuse/side opposite to ∠θ = AC/BC = 13/5