Question

Let $a$ be the hypotenuse and $b$ be the base of a right-angled triangle. If the area of the triangle is equal to $\frac{ab}{4}$ square units then find the angle between the base and the hypotenuse.

Answer 1

Let the angle between base and hypotenuse be $\theta .$

In $△XYZ,$

$\sin \theta =\frac{YZ}{XZ}$

$\Rightarrow YZ=XZ\times \sin \theta$

$\Rightarrow YZ=a\sin \theta$

Now,

$\begin{array}{cc}ar\left(△XYZ\right)& =\frac{1}{2}\times YZ\times XY\\ \frac{ab}{4}& =\frac{1}{2}\times a\sin \theta \times b\\ \frac{1}{2}\times \frac{ab}{2}& =\frac{ab}{2}\times \sin \theta \\ \sin \theta & =\frac{1}{2}\\ \theta & ={30}^{\circ }\end{array}$