Question

In an acute angled ABC, if $ \tan (A + B - C) = 1 $ and $ \sec (B + C - A) = 2 $ , find the value of A, B and C.

Answer 1

Hint: By using the trigonometric table, we will form equations and further simplify the equations A, B and C values can be calculated. In RHS, we can replace 1 as $ tan {45^o} $. From there we will find equations in the form of A, B, and C. On solving these equations, we will get the value of A,B and C.

Complete step-by-step answer:
According to the question, we have
 $ \Rightarrow \tan (A + B + C) = 1 $ and $ \sec (B + C - A) = 2 $
 $ \Rightarrow \tan (A + B + C) = \tan {45^o}[\because \tan {45^o} = 1] $
and $ \sec (B + C - A) = \sec {60^o}[\because \sec {60^o} = 2] $
 $ \Rightarrow A + B - C = {45^o} \to (1) $
 $ B + C - A = {60^o} \to (2) $
Adding equation (1) and (2), we get
 $
   \Rightarrow (A + B - C) + (B + C - A) = {45^o} + {60^o} \\
   \Rightarrow 2B = {105^o} \\
   \Rightarrow B = 52{\dfrac{1}{2}^o} \\
 $
Now as per the calculation above, putting $ B = 52{\dfrac{1}{2}^o} $ in equation (2), we get
 $
   \Rightarrow 52{\dfrac{1}{2}^o} + C - A = 60 \\
   \Rightarrow C - A = 7{\dfrac{1}{2}^o} \to (3) \\
  $
As we know, the sum of angles in a triangle is $ {180^o} $ .Therefore, we get
 $
   \Rightarrow A + B + C = {180^o} \\
   \Rightarrow A + C = {180^o} - 52{\dfrac{1}{2}^o}[\because B = 52{\dfrac{1}{2}^o}] \\
   \Rightarrow A + C = 127{\dfrac{1}{2}^o} \to (4) \\
 $
Adding equation (3) and (4), we get
 $
   \Rightarrow 2C = {135^o} \\
   \Rightarrow C = 67{\dfrac{1}{2}^o} \\
 $
Substituting the obtained value in equation (3), we get
 $
   \Rightarrow 67{\dfrac{1}{2}^o} - A = 7{\dfrac{1}{2}^o} \\
   \Rightarrow A = {60^o} \\
 $
Hence, $ A = {60^o},B = 52{\dfrac{1}{2}^o},C = 67{\dfrac{1}{2}^o} $ .

Note: For this type of question, read the question carefully. Make equations separately and relate them to each other. You will find it useful. Remember basic trigonometric identities and values.