In an acute angled triangle ABC, if tan (A+B-C) = 1 and sec (B+C-A) =

1.	In an acute angled triangle ABC, if tan (A+B-C) = 1 and sec (B+C-A) = 2, find the values of A, B and C.
Answer» Tan (A+B-C) = 1 Tan 45° = 1 Hence, A+B-C = 45° ........(i) Sec (B+C-A) = 2 Sec 60° = 2 Hence, B+C-A = 60° .......(ii) Adding equation (i) and (ii), we get, A+B-C + B+C-A = 45° + 60° 2B = 105° B = 52.5° Hence, A+B-C = 45° A+52.5° -C = 45° A-C = 45° - 52.5° A-C = -7.5° ............(iii) Now, A+B+C = 180° Hence, A + 52.5° + C = 180° (Angles of the triangle) A + C = 180° - 52.5° A + C = 127.5° .........(iv) Now, Adding equation (iii) and (iv), we get, A-C + A + C = -7.5° + 127.5° 2A = 120° A = 120°/2 A = 60° C = 127.5° - 60° C = 67.5° Hence, A = 60°, B = 52.5° and C = 67.5°.

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

Answer»

Tan (A+B-C) = 1

Tan 45° = 1

Hence, A+B-C = 45° ........(i)

Sec (B+C-A) = 2

Sec 60° = 2

Hence, B+C-A = 60° .......(ii)

Adding equation (i) and (ii), we get,

A+B-C + B+C-A = 45° + 60°

2B = 105°

B = 52.5°

Hence, A+B-C = 45°

A+52.5° -C = 45°

A-C = 45° - 52.5°

A-C = -7.5° ............(iii)

Now,

A+B+C = 180°

Hence,

A + 52.5° + C = 180° (Angles of the triangle)

A + C = 180° - 52.5°

A + C = 127.5° .........(iv)

Now, Adding equation (iii) and (iv), we get,

A-C + A + C = -7.5° + 127.5°

2A = 120°

A = 120°/2

A = 60°

C = 127.5° - 60°

C = 67.5°

Hence,

A = 60°, B = 52.5° and C = 67.5°.

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