{"id":11677,"date":"2022-08-14T01:48:59","date_gmt":"2022-08-13T20:18:59","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11677"},"modified":"2022-08-14T01:48:31","modified_gmt":"2022-08-13T20:18:31","slug":"in-triangle-pqr-right-angled-at-q-pr-qr-25-cm-and-pq-5-cm-determine-the-values-of-sin-p-cos-p-and-tan-p","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-triangle-pqr-right-angled-at-q-pr-qr-25-cm-and-pq-5-cm-determine-the-values-of-sin-p-cos-p-and-tan-p\/","title":{"rendered":"In \\(\\triangle\\) PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P."},"content":{"rendered":"
We have,<\/p>\n
PQ = 5 cm<\/p>\n
PR + QR = 25 cm\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(1)<\/p>\n
In triangle PQR, By Pythagoras Theorem,<\/p>\n
\\({PR}^2\\) = \\({PQ}^2\\) + \\({QR}^2\\)<\/p>\n
\\(\\implies\\)\u00a0 \\({PQ}^2\\) = \\({PR}^2\\) – \\({QR}^2\\)<\/p>\n
\\(\\implies\\)\u00a0 \\({PQ}^2\\) = (PR + QR)(PR – QR)<\/p>\n
\\(\\implies\\)\u00a0 \\(5^2\\) = (PR – QR). 25<\/p>\n
\\(\\implies\\)\u00a0 PR – QR = 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………..(2)<\/p>\n
Solving (1) and (2), we have<\/p>\n
PR = 13 cm and QR = 12 cm<\/p>\n
sin P = \\(QR\\over PR\\) = \\(12\\over 13\\)<\/strong><\/p>\n cos P = \\(PQ\\over PR\\) = \\(5\\over 13\\)<\/strong><\/p>\n tan P = \\(QR\\over PQ\\) = \\(12\\over 5\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : We have, PQ = 5 cm PR + QR = 25 cm\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(1) In triangle PQR, By Pythagoras Theorem, \\({PR}^2\\) = \\({PQ}^2\\) + \\({QR}^2\\) \\(\\implies\\)\u00a0 \\({PQ}^2\\) = \\({PR}^2\\) – \\({QR}^2\\) \\(\\implies\\)\u00a0 \\({PQ}^2\\) = (PR + QR)(PR – QR) \\(\\implies\\)\u00a0 \\(5^2\\) = (PR – QR). 25 …<\/p>\n