Check sibling questions

Ex 8.1, 10 - In PQR, PR + QR = 25 cm and PQ = 5 cm. - Finding ratios when sides are given

Ex 8.1, 10 - Chapter 8 Class 10 Introduction to Trignometry - Part 2
Ex 8.1, 10 - Chapter 8 Class 10 Introduction to Trignometry - Part 3
Ex 8.1, 10 - Chapter 8 Class 10 Introduction to Trignometry - Part 4

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Ex 8.1, 10 In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. Given PR + QR = 25 cm Let QR = x PR + QR = 25 cm PR = 25 – QR PR = 25 – x In right triangle PQR, Using pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 PR2 = PQ2 + QR2 (25 – x)2 = 52 + x2 (25)2 + 𝑥2 − 2×25×𝑥=25+𝑥2 625 + x2 – 50x = 25 + x2 625 + x2 – 50x – 25 – x2 = 0 x2 – x2 – 50x + 625 – 25 = 0 − 50𝑥 + 600 = 0 − 50𝑥 = −600 x = (− 600)/(− 50) x = 12 Hence, QR = x = 12 cm PR = 25 – x = 25 – 12 = 13 cm And tan P = (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑃 )/(𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑃) tan P = 𝑄𝑅/𝑃𝑄 tan P = 12/5 tan P = 12/5 Alternatively, tan P = 𝑠𝑖𝑛⁡P/𝑐𝑜𝑠⁡P tan P = (12/13)/(5/13) tan P = 12/5

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.