Question

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

Answer 1

Answer:

Step-by-step explanation:

Tan ( A + b - c) = 1

tan ( a+ b - c) = tan 45°

a+b+-c = 45°. {1}

By angle sum property

a +b+ c = 180°. { 2}

also, sec( b +c - a) = 2

b + c - a = 60°. { 3}

from eqn 1 we get

a + b = 45 + C

put this value in eqn 2

45+c + c = 180

2c = 180 - 45

c = 135/2 = 67.5 °

put this value in eqn 1

a +b = 45+67.5

a+ b = 112.5 (4)

also in eqn 3

B + 67.5 - A = 60

-A + B = - 7.5

A - B = 7.5 (5)

eliminating (4) from (5)

A = 105/2 ° = 52.5°

B = 45°

This will help you