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UPSC CDS 1 2022 Math Paper

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Question : 78 of 100

Marks: +1, -0

If

A, B, C

are acute angles and

\sin (B+C-A)=\cos (C+A-B)=\tan (A+B-C)=1

(A+B+C)

equal to?

$90^{\circ}$
$120^{\circ}$
$135^{\circ}$
$150^{\circ}$

Solution: 👈: Video Solution

Given:

A, B, C

are acute angles

\sin (B+C-A)=\cos (C+A-B)=\tan (A+B-C)=1

Concept used:

0°	30°	45°	60°	90°
sin	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
cos	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0
tan	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$
cosec	$\infty$	$2$	$\sqrt{2}$	$\frac{2}{\sqrt{3}}$	$1$
sec	$1$	$\frac{2}{\sqrt{3}}$	$\sqrt{2}$	$2$	$\infty$
cot	$\infty$	$\sqrt{3}$	$1$	$\frac{1}{\sqrt{3}}$	$0$

\sin (B+C-A)=1

So,

\sin (B+C-A)=\sin 90^{\circ}

\Rightarrow (B+C-A)=90^{\circ} \;\; \cdots (i)

\cos (C+A-B)=1

So,

\cos (C+A-B)=\cos 0^{\circ}

\Rightarrow (C+A-B)=0^{\circ} \;\; \cdots

(ii)

\tan (A+B-C)=1

So,

\tan (A+B-C)=\tan 45^{\circ}

\Rightarrow (A+B-C)=45^{\circ}

By adding (i), (ii), and (iii) we get

B+C-A+C+A-B+A+B-C=90^{\circ}+0^{\circ}+45^{\circ}

\Rightarrow A+B+C=135^{\circ}

\therefore A+B+C

is equal to

135^{\circ}

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