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Q6:
If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.

Solution :

Given: $\angle A$ and $\angle B$ are acute angles in a right-angled triangle, such that $\cos A = \cos B$.

To Prove: $\angle A = \angle B$.

A C B AC BC AB

Step 1: Defining the Trigonometric Ratios
Consider a right-angled triangle $ABC$ where $\angle C = 90^\circ$. Since $\angle A$ and $\angle B$ are acute angles, they are the other two angles of the triangle.

By the definition of the cosine ratio in a right-angled triangle:

$\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$

Step 2: Expressing $\cos A$ and $\cos B$
For $\angle A$:
The side adjacent to $\angle A$ is $AC$.
The hypotenuse is $AB$.
Therefore, $\cos A = \frac{AC}{AB}$ --- (Equation 1)

For $\angle B$:
The side adjacent to $\angle B$ is $BC$.
The hypotenuse is $AB$.
Therefore, $\cos B = \frac{BC}{AB}$ --- (Equation 2)

Step 3: Equating the Ratios
Given that $\cos A = \cos B$, we substitute the expressions from Equation 1 and Equation 2:

$\frac{AC}{AB} = \frac{BC}{AB}$

Step 4: Algebraic Simplification
Multiply both sides of the equation by $AB$ (since $AB$ is a side length of a triangle, $AB \neq 0$):

$AC = BC$

Step 5: Applying Geometric Theorems
In $\triangle ABC$, we have shown that side $AC = BC$.
According to the Isosceles Triangle Theorem, angles opposite to equal sides of a triangle are equal.

Since $AC$ is the side opposite to $\angle B$ and $BC$ is the side opposite to $\angle A$:

$\angle A = \angle B$

Final Answer: Since the sides opposite to the equal angles are equal, it is proven that $\angle A = \angle B$.


More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.1


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