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

According to the question,

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