Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Let $\mathrm{ED}=h,\angle \mathrm{EAB}=\alpha$ $\therefore \angle \mathrm{EBD}=2\alpha ,\angle \mathrm{ECD}=3\alpha$ Now, $\angle \mathrm{DBE}=\angle \mathrm{EAB}+\angle \mathrm{BEA}$ $\Rightarrow 2\alpha =\alpha +\angle \mathrm{BEA}$ $\Rightarrow \angle \mathrm{BEA}=\alpha =\angle \mathrm{EAB}$ $\Rightarrow \mathrm{AB}=\mathrm{EB}=a$ Similarly, $\angle \mathrm{EBC}=\alpha$ $\mathrm{In}\Delta \mathrm{EBC},\frac{\mathrm{BC}}{\sin \alpha }=\frac{\mathrm{EB}}{\sin \left(180^{\circ }-3\alpha \right)}$ $\Rightarrow \frac{b}{\sin \alpha }=\frac{a}{\sin 3\alpha }\Rightarrow \frac{a}{b}=\frac{\sin 3\alpha }{\sin 3\alpha }$ $\Rightarrow \frac{a}{b}=\frac{3\sin \alpha -4{\sin }^3\alpha }{\sin \alpha }=3-4{\sin }^2\alpha$ $\Rightarrow 4{\sin }^2\alpha =3-\frac{a}{b}=\frac{3b-a}{b}$ $\Rightarrow \sin \alpha =\sqrt{\frac{3b-a}{4b}}$ In $\Delta \mathrm{EBD},\sin 2\alpha =\frac{ED}{EB}$ $\Rightarrow \mathrm{ED}=\mathrm{a}\cdot 2\sin \alpha \cdot \cos \alpha$ $\Rightarrow h=2a\sqrt{\frac{3b-a}{4b}},\sqrt{1-\frac{3b-a}{4b}}$ $=2a\sqrt{\frac{3b-a}{4b}}\sqrt{\frac{b-a}{4b}}$ $=\frac{a}{2b}\sqrt{(a+b)(3b-a)}$