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

Solution :

Let us consider two right triangles PQA and RSB in which cos A = cos B (see figure).

We have :       cos A = $$QA\over PA$$

and         cos B = $$SB\over RB$$

Thus, it is given that

$$QA\over PA$$ = $$SB\over RB$$

So, $$QA\over SB$$ = $$PA\over RB$$ = k (say)

Now, By Pythagoras Theorem,

PQ = $$\sqrt{{PA}^2 – {QA}^2}$$

and RS = $$\sqrt{{RB}^2 – {SB}^2}$$

So, $$PQ\over RS$$ = $$\sqrt{{PA}^2 – {QA}^2}\over \sqrt{{RB}^2 – {SB}^2}$$

or  $$PQ\over RS$$ = $$k\sqrt{{PA}^2 – {QA}^2}\over \sqrt{{RB}^2 – {SB}^2}$$ = k

Therefore, from (1) and (2), we have :

$$QA\over SB$$ = $$PA\over RB$$ = $$PQ\over RS$$

By SSS similarity,

$$\triangle$$ PQA ~ $$\triangle$$ RSB

Therefore,  $$\angle$$ A = $$\angle$$ B              (Corresponding Angles)