Ex 8.1

Chapter 8 Class 10 Introduction to Trignometry
Serial order wise

### Transcript

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 Thus, 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 sin P = (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑃 )/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝑄𝑅/𝑃𝑅 = 𝟏𝟐/𝟏𝟑 cos P = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑃 )/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝑃𝑄/𝑃𝑅 = 𝟓/𝟏𝟑 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

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.