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NCERT Solutions 10th Maths Chapter 8 Introduction to Trigonometry Exercise 8.3
NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.3 we will restart our exploration of the world of Introduction to Trigonometry. Thus, we are providing you Chapter 8 Introduction to Trigonometry NCERT Class 10 Maths Solutions that will help in achieving more marks. You don't have to wander and waste your precious time in finding best CBSE NCERT Solutions.
Exercise 8.3 1. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A. Solution: We have tan 2A = cot (A – 18°) ⇒ cot (90° – 2A) = cot (A – 18°) [ ∵ cot (90° – θ) = tan θ] ⇒ 90° – 2A = A – 18° ⇒ 2A + A = 90° + 18° ⇒ 3A = 108° ⇒ A = \frac { 108 }{ 3 } = 36°. 2. If tan A = cot B, prove that A + B = 90°. Solution: We have: tan A = cot B ⇒ cot (90° – A) = cot B ⇒ 90° – A = B ⇒ A + B = 90° Hence, Proved. 3.If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A. Solution: We have sec 4A = cosec (A – 20°) ⇒ cosec (90° – 4A) = cosec (A – 20°) [∵ cosec (90° – θ) = sec θ] ⇒ 90° – 4A = A – 20° ⇒ 90° + 20° = 5A ⇒ 110° = 5A ⇒ \frac { 110 }{ 5 } = A ⇒ A = 22° 4.If A, B and C are interior angles of a triangle ABC, then show that: sin (\frac { B+C }{ 2 }) = cos \frac { A }{ 2 } Solution:Post your comments
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