Question

In the adjoining figure, $\angle B={90}^0,\angle BAC={\theta }^0,BC=CD=4cm$ and $AD=10cm$. If $\sin \theta =\frac{2}{\sqrt{m}}$. Find $m$.

Answer 1

In $△ABD$,

By hypotenuese theorem,

${AD}^2={AB}^2+{BD}^2$

${\left(10\right)}^2={AB}^2+{\left(8\right)}^2\left(\text{From fig.}\right)$

$\Rightarrow {AB}^2=100-64$

$\Rightarrow AB=\sqrt{36}=6$

Now, in $△ABC$,

By hypotenuse theorem,

${AC}^2={AB}^2+{BC}^2$

${AC}^2=6^2+4^2$

$\Rightarrow {AC}^2=36+16$

$\Rightarrow AC=\sqrt{52}=2\sqrt{13}$

Now, again in $△ABC$,

$\sin \theta =\frac{BC}{AC}(\because \sin \theta =\frac{P}{h})$

$\Rightarrow \sin \theta =\frac{4}{2\sqrt{13}}$

$\Rightarrow \sin \theta =\frac{2}{\sqrt{13}}⟶\left(1\right)$

Given that,

$\sin \theta =\frac{2}{\sqrt{m}}$

From ${eq}^n\left(1\right)$, we have

$m=13$

Hence, the correct answer is 13.