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).triangles

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)

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