ABCD is a trapezium such that AB and CD are parallel and BC⊥CD. If ∠ADB = θ, BC = p and CD = q, then AB is equal to :

Options

A)

B)

C)

D)

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$$\begin{gathered} \mathrm{In} ∆\mathrm{BCD} \\ \Rightarrow \mathrm{BC}=\mathrm{p} \\ \Rightarrow \mathrm{CD}=\mathrm{q} \\ \mathrm{In} \mathrm{right} \mathrm{angled} \mathrm{triangle} ∆\mathrm{BCD} \\ \Rightarrow \mathrm{BD}=\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2} \\ \mathrm{Let} \angle \mathrm{BDC}=\mathrm{\alpha } \\ \mathrm{As} \mathrm{BC}\perp \mathrm{CD} \\ \Rightarrow \mathrm{CD}\parallel \mathrm{AB} \\ \Rightarrow \angle \mathrm{DBA}=\angle \mathrm{BDC}=\mathrm{\alpha } \left[\mathrm{Alternate} \mathrm{angles}\right] \\ \mathrm{In} ∆\mathrm{ABD} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{is} \mathrm{\pi } \\ \Rightarrow \mathrm{\alpha }+\mathrm{\theta }+\angle \mathrm{A}=\mathrm{\pi } \\ \Rightarrow \angle \mathrm{A}=\mathrm{\pi }-\left(\mathrm{\alpha }+\mathrm{\theta }\right) \\ \mathrm{In} ∆\mathrm{ABD}, \mathrm{use} \mathrm{sine} \mathrm{rule} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{sin\theta }}=\frac{\mathrm{BD}}{\sin \left\{\mathrm{\pi }-\left(\mathrm{\alpha }+\mathrm{\theta }\right)\right\}} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{sin\theta }}=\frac{\mathrm{BD}}{\sin \left(\mathrm{\alpha }+\mathrm{\theta }\right)} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{sin\theta }}=\frac{\mathrm{BD}}{\mathrm{sin\theta cos\alpha }+\mathrm{cos\theta sin\alpha }} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{sin\theta }}=\frac{\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}}{\mathrm{sin\theta }.{\displaystyle \frac{\mathrm{q}}{\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}}}+\mathrm{cos\theta }.{\displaystyle \frac{\mathrm{p}}{\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}}}} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{sin\theta }}=\frac{\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}.\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2}}{\mathrm{pcos\theta }+\mathrm{qsin\theta }} \\ \mathbf{\Rightarrow }\mathbf{AB}\mathbf{=}\frac{\left({\mathbf{p}}^{\mathbf{2}}\mathbf{+}{\mathbf{q}}^{\mathbf{2}}\right)\mathbf{sin\theta }}{\mathbf{pcos\theta }\mathbf{+}\mathbf{qsin\theta }} \end{gathered}$$

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