# If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

**Solution:**

Using the basic trigonometric ratios, we can solve this problem.

In the right-angled triangle ABC as shown below, ∠A and ∠B are acute angles and ∠C is right angle.

cos A = side adjacent to ∠A / hypotenuse = AC/AB

cos B = side adjacent to ∠B / hypotenuse = BC/AB

Given that cos A = cos B

Therefore, AC/AB = BC/AB

AC = BC

Hence, ∠A = ∠B (angles opposite to equal sides of a triangle are equal.)

Let's look into an alternative approach to solve the question.

Let us consider a triangle ABC in which CO ⊥ AB.

It is given that cos A = cos B

AO/AC = BO/BC

AO/BO = AC/BC

Let AO/BO = AC/BC = k

AO = k × BO ...(i)

AC = k × BC ...(ii)

By applying Pythagoras theorem in ΔCAO and ΔCBO, we get

AC^{2} = AO^{2} + CO^{2} (from ΔCAO)

CO^{2} = AC^{2} - AO^{2} ...(iii)

BC^{2} = BO^{2} + CO^{2} (from ΔCBO)

CO^{2} = BC^{2} - BO^{2} ...(iv)

From equation (iii) and equation (iv), we get

AC^{2} - AO^{2} = BC^{2} - BO^{2}

(kBC)^{2} - (kBO)^{2} = BC^{2} - BO^{2} [From equation (i) and (ii)]

k^{2}BC^{2} - k^{2}BO^{2} = BC^{2} - BO^{2}

k^{2} (BC^{2} - BO^{2}) = BC^{2} - BO^{2}

k^{2} = (BC^{2} - BO^{2})/(BC^{2} - BO^{2}) = 1

k = 1

Putting this value in equation (ii) we obtain,

AC = BC

Thus, ∠A = ∠B (angles opposite to equal sides of triangle are equal.)

**Video Solution:**

## If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

### Maths NCERT Solutions Class 10 - Chapter 8 Exercise 8.1 Question 6:

**Summary:**

If ∠A and ∠B are acute angles such that cos A = cos B we have proved that ∠A = ∠B since AC = BC and angles opposite to equal sides of a triangle are equal.