Question

If $\sin {90}^{\circ }-(A+B)=\cos x=\cos y,$

find the value of $x$ and $y$; if $y$ is the angle $C$ of triangle $ABC.$ (In triangle $ABC,\angle A+\angle B={90}^{\circ }$)

• A. x = A+B; y=C
• B. x=A+B; y =
• C. both (a) and (b)
• D. none of these

Answer 1

The correct option is C

both (a) and (b)

$\sin {90}^{\circ }-(A+B)=\cos x$

$\Rightarrow \sin {90}^{\circ }-(A+B)=\sin ({90}^{\circ }-x)$

Thus, $x=A+B$

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

$\Rightarrow \sin {90}^{\circ }-(A+B)=\sin ({90}^{\circ }$ - y)\)

Thus, $y=A+B$

Now, In triangle $ABC,\angle A+\angle B+\angle C={180}^{\circ }$

and given, $\angle A+\angle B={90}^{\circ }$ and $y=\angle C$ of triangle ABC,

Hence, $y=\angle C={90}^{\circ }$

Also, $x=\angle A+\angle B=y={90}^{\circ }$