One of the angles of a right-angled triangle is 15°, and the hypotenuse is 1 m. The area of the triangle (in sq. cm.) is

Answer: Option B

Solution

Answer: Option B

Solution:

According to the question,

$$\begin{array}{rl}& \sin 15^{\circ }=\frac{P}{H}=\frac{AB}{1}\\ & AB=\sin 15^{\circ }\\ & \cos 15^{\circ }=\frac{B}{H}=\frac{BC}{1}\\ & BC=\cos 15^{\circ }\\ & \text{Area of }\Delta ABC=\frac{1}{2}\times AB\times BC\\ & =\frac{1}{2}\times \sin 15^{\circ }\times \cos 15^{\circ }\\ & =\frac{1}{2\times 2}\times 2\sin 15^{\circ }\times \cos 15^{\circ }\\ & =\frac{1}{4}\times \sin 2\times 15^{\circ }\left[\because \sin 2\theta =2\sin \theta \cos \theta \right]\\ & =\frac{1}{4}\times \sin 30^{\circ }\\ & =\frac{1}{4}\times \frac{1}{2}\\ & =\frac{1}{8}{\text{m}}^2\\ & =\frac{1}{8}\times 100\times 100=1250\text{ c}{\text{m}}^2\end{array}$$