{"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":"

Solution :<\/h2>\n

We have,\"triangle\"<\/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

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.<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,60],"tags":[],"yoast_head":"\nIn \\(\\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. - Mathemerize<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"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\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"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. - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"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 … 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. 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Determine the values of sin P, cos P and tan P. 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