Question

The value of $\sin$ ${45}^{\circ }$ is

• A. 1/2
• B. √3
• C. √3/2
• D. 1/ √2

Answer 1

The correct option is D

1/ √2

Consider the right isosceles triangle $\Delta ABC$ in figure having base and perpendicular AB and BC respectively of 1 unit length each.
$\therefore AC=\sqrt{{AB}^2+{BC}^2}=\sqrt{1^2+1^2}=\sqrt{2}Sincethe\Delta isisosceles,\angle CAB=\angle ACBalso\angle CAB+\angle ACB+\angle ABC={180}^{\circ }\Rightarrow \angle CAB+\angle ACB+{90}^{\circ }={180}^{\circ }\Rightarrow \angle CAB=\angle ACB={45}^{\circ }\therefore sin\angle CAB=sin{45}^{\circ }=\frac{BC}{AC}\Rightarrow sin{45}^{\circ }=\frac{1}{\sqrt{2}}$