Last updated at June 23, 2017 by Teachoo

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 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 = 𝑠𝑖𝑛P/𝑐𝑜𝑠P tan P = (12/13)/(5/13) tan P = 12/5

Example 2
Important

Example 6 Important

Example 7 Important

Example 11 Important

Example 15 Important

Ex 8.1, 5 Important

Ex 8.1, 10 Important You are here

Ex 8.1, 9 Important

Ex 8.3, 5 Important

Ex 8.1, 11 Important

Ex 8.3, 6 Important

Ex 8.3, 7 Important

Ex 8.4, 1 Important

Ex 8.4, 2 Important

Ex 8.4, 5 Important

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.