In an acute angle triangle ABC, $sin\ (A + B - C) = \frac{1}{2}$, $cot\ (A - B + C) = 0$ and $cos (B + C - A) =\frac{1}{2}$. What are the values of A, B, and C?

Given:

In an acute angle triangle ABC, $sin\ (A + B - C) = \frac{1}{2}$, $cot\ (A - B + C) = 0$ and $cos (B + C - A) =\frac{1}{2}$.

To do:

We have to find the values of A, B and C.

Solution:

We know that,

Sum of the angles in a triangle is $180^o$.

$A+B+C=180^o$......(i)

$sin\ 30^o=\frac{1}{2}$

$cos\ 60^o=\frac{1}{2}$

Therefore,

$sin\ (A+B-C) = sin\ 30^o$

$\Rightarrow A+B-C = 30^o$.......(ii)

$cos\ (B+C-A) = cos\ 60^o$

$B+C-A = 60^o$.....(iii)

Adding equations (ii) and (iii), we get,  

$A+B-C+B+C-A = 30^o+60^o = 90^o$

$2B=90^o$

$B=45^o$

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

$A+B+C+A+B-C = 180^o+30^o = 210^o$

$2(A+B)=210^o$

$A+B=105^o$

$\Rightarrow A+45^o=105^o$

$A=105^o-45^o$

$A=60^o$

Substituting the values of angles A and B in (i), we get,

$60^o+45^o+C=180^o$

$C=180^o-105^o$

$C=75^o$

Therefore, $A=60^o, B=45^o$ and $C=75^o$.

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Updated on: 10-Oct-2022

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