Question : If A, B, and C are the acute angles, such that $\tan (A+B-C)=\frac{1}{\sqrt{3}}, ~\cos (B+C-A)=\frac{1}{2},$ and $\sin (C+A-B)=\frac{1}{\sqrt{2}}$. The value of A + B + C will be equal to:
Option 1: 90°
Option 2: 110°
Option 3: 170°
Option 4: 135°
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Correct Answer: 135°
Solution :
Given, $\tan (A+B-C)=\frac{1}{\sqrt{3}}, ~\cos (B+C-A)=\frac{1}{2},$ and $\sin (C+A-B)=\frac{1}{\sqrt{2}}$
A, B, and C are acute angles
For $\tan (A+B-C)=\frac{1}{\sqrt{3}}=\tan30°$
$⇒A+B-C =30°$ -----------(1)
For $\cos (B+C-A)=\frac{1}{2}=\cos60°$
$⇒B+C-A=60°$ ------------(2)
And for $\sin (C+A-B)=\frac{1}{\sqrt{2}}=\sin45°$
$⇒C+A-B =45°$ --------(3)
Adding equations 1, 2, and 3, we get,
⇒ $A+B+C =30°+60°+45° = 135°$
Hence, the correct answer is 135°.
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