Question

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 1

from question,
tan(theta¹)=1..So,theta¹=45°
sec(theta²)=3..so,theta²=60°

A+B-C=45° eq1
B+C-A=60° eq2
ALSO, A+B+C=180° eq3
NOW, ADDING 1 &2
2(B)=105
B=52.5°
ADDING 1&3
2(A+B)=225°
A=112.5°-52.5°
A=60°
now adding 2&3
2(B+C)=240
C=120°-52.5°
C=67.5°

Answer 2

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"Good question".

Here is probably the perfect answer..

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