question_answer

In \[\Delta ABC\], if\[AB=BC\], \[\angle B=90{}^\circ \], then the value sin A is

A) \[\frac{1}{\sqrt{2}}\]

B) \[\frac{\sqrt{3}}{2}\]

C) 0

D) \[\frac{1}{2}\]

Correct Answer: A

Solution :

Given, in \[\Delta ABC\], \[\angle B=90{}^\circ\]and AB=BC

\[\therefore \,\,\angle ACB=\angle BAC\]

[\[\because\] angles opposite to two equal sides are equal]

In \[\Delta ABC,\]\[\angle ABC+\angle ACB+\angle BAC=180{}^\circ\]

Let \[\angle ACB=\angle BAC=x\]

\[\therefore \,\,\,90{}^\circ \,+x+x=180{}^\circ\]

[\[\therefore\]sum of all angles of a triangle is \[180{}^\circ\]]

\[\Rightarrow \,\,\,\,2x=90{}^\circ \Rightarrow \,x=45{}^\circ\]

\[\therefore \,\,\,\angle ACB=\angle BAC=45{}^\circ\]

(i) \[\sin A=\sin 45{}^\circ =\frac{1}{\sqrt{2}}\]  \[\left[ \because \,\sin \,45{}^\circ =\frac{1}{\sqrt{2}} \right]\]

(ii) \[\operatorname{cosA}=cos45{}^\circ =\frac{1}{\sqrt{2}}\]   \[\left[ \because \,\,\cos \,45{}^\circ =\frac{1}{\sqrt{2}} \right]\]