![\"triangles\"](\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/08\/trigonometry8-1-6.jpeg?resize=285%2C192&ssl=1\")
<\/p>\nLet us consider two right triangles PQA and RSB in which cos A = cos B (see figure).<\/p>\n
We have :\u00a0 \u00a0 \u00a0 \u00a0cos A = \\(QA\\over PA\\)<\/p>\n
and\u00a0 \u00a0 \u00a0 \u00a0 \u00a0cos B = \\(SB\\over RB\\)<\/p>\n
Thus, it is given that<\/p>\n
\\(QA\\over PA\\) = \\(SB\\over RB\\)<\/p>\n
So, \\(QA\\over SB\\) = \\(PA\\over RB\\) = k (say)<\/p>\n
Now, By Pythagoras Theorem,<\/p>\n
PQ = \\(\\sqrt{{PA}^2 – {QA}^2}\\)<\/p>\n
and RS = \\(\\sqrt{{RB}^2 – {SB}^2}\\)<\/p>\n
So, \\(PQ\\over RS\\) = \\(\\sqrt{{PA}^2 – {QA}^2}\\over \\sqrt{{RB}^2 – {SB}^2}\\)<\/p>\n
or\u00a0 \\(PQ\\over RS\\) = \\(k\\sqrt{{PA}^2 – {QA}^2}\\over \\sqrt{{RB}^2 – {SB}^2}\\) = k<\/p>\n
Therefore, from (1) and (2), we have :<\/p>\n
\\(QA\\over SB\\) = \\(PA\\over RB\\) = \\(PQ\\over RS\\)<\/p>\n
By SSS similarity,<\/p>\n
\\(\\triangle\\) PQA ~ \\(\\triangle\\) RSB<\/p>\n
Therefore,\u00a0 \\(\\angle\\) A = \\(\\angle\\) B\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (Corresponding Angles)<\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : Let us consider two right triangles PQA and RSB in which cos A = cos B (see figure). We have :\u00a0 \u00a0 \u00a0 \u00a0cos A = \\(QA\\over PA\\) and\u00a0 \u00a0 \u00a0 \u00a0 \u00a0cos B = \\(SB\\over RB\\) Thus, it is given that \\(QA\\over PA\\) = \\(SB\\over RB\\) So, \\(QA\\over SB\\) = \\(PA\\over RB\\) …<\/p>\n