Question

In an acute angled triangle ABC if, tan (A+ B- C)=√3 and sec (B+C - A)=1 Find angleA, angleB and angleC.

Answer 1

Answer:

A = 90°

B = 30°

C = 60°

Step-by-step explanation:

tan ( A + B - C ) = ✓3

Also, tan 60° = ✓3

Hence, A + B - C = 60° ------- ( 1 )

Next, sec ( B + C - A ) = 1

However, sec 0° = 1

Hence, B + C - A = 0°

Thus, B + C = A ----- ( 2 )

By ( 1 ) + ( 2 ),

A + 2B = 60° + A

2B = 60°

B = 30°

Therefore, in ( 1 ),

A + 30° - C = 60°

A - C = 30° ----- ( 3 )

Now, by Angle Sum Property,

A + B + C = 180°

A + 30° + C = 180°

A + C = 150° ----- ( 4 )

By ( 3 ) + ( 4 ),

2A = 180°

A = 90°

Hence, in ( 3 ),

90° - C = 30°

C = 60°

Thus,

A = 90°

B = 30°

C = 60°