Properties of Triangles 1 Question 4
4. $A B C D$ is a trapezium such that $A B$ and $C D$ are parallel and $B C \perp C D$, if $\angle A D B=\theta, B C=p$ and $C D=q$, then $A B$ is equal to
(2013 Main)
(a) $\frac{\left(p^{2}+q^{2}\right) \sin \theta}{p \cos \theta+q \sin \theta}$
(b) $\frac{p^{2}+q^{2} \cos \theta}{p \cos \theta+q \sin \theta}$
(c) $\frac{p^{2}+q^{2}}{p^{2} \cos \theta+q^{2} \sin \theta}$
(d) $\frac{\left(p^{2}+q^{2}\right) \sin \theta}{(p \cos \theta+q \sin \theta)^{2}}$
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Answer:
Correct Answer: 4. (a)
Solution:
- Applying sine rule in $\triangle A B D$,
$$ \begin{aligned} \frac{A B}{\sin \theta} & =\frac{\sqrt{p^{2}+q^{2}}}{\sin {\pi-(\theta+\alpha)}} \\ \Rightarrow \quad \frac{A B}{\sin \theta} & =\frac{\sqrt{p^{2}+q^{2}}}{\sin (\theta+\alpha)} \\ \Rightarrow \quad A B & =\frac{\sqrt{p^{2}+q^{2}} \sin \theta}{\sin \theta \cos \alpha+\cos \theta \sin \alpha} \quad \because \cos \alpha=\frac{q}{\sqrt{p^{2}+q^{2}}} \end{aligned} $$
$$ =\frac{\left(p^{2}+q^{2}\right) \sin \theta}{p \cos \theta+q \sin \theta} $$
and $\sin \alpha=\frac{p}{\sqrt{p^{2}+q^{2}}}$
Alternate Solution
Let $\quad A B=x$
In $\triangle D A M, \tan (\pi-\theta-\alpha)=\frac{p}{x-q}$
$$ \Rightarrow \quad \tan (\theta+\alpha)=\frac{p}{q-x} $$
$$ \begin{aligned} \Rightarrow & & q-x & =p \cot (\theta+\alpha) \\ \Rightarrow & & x & =q-p \cot (\theta+\alpha) \end{aligned} $$
$$ =q-p \frac{\cot \theta \cot \alpha-1}{\cot \alpha+\cot \theta} \quad \because \cot \alpha=\frac{q}{p} $$
$$ =q-p \frac{\frac{q}{p} \cot \theta-1}{\frac{q}{p}+\cot \theta}=q-p \frac{q \cot \theta-p}{q+p \cot \theta} $$
$$ =q-p \frac{q \cos \theta-p \sin \theta}{q \sin \theta+p \cos \theta} $$
$$ \begin{array}{ll} \Rightarrow & x=\frac{q^{2} \sin \theta+p q \cos \theta-p q \cos \theta+p^{2} \sin \theta}{p \cos \theta+q \sin \theta} \\ \Rightarrow & A B=\frac{\left(p^{2}+q^{2}\right) \sin \theta}{p \cos \theta+q \sin \theta} \end{array} $$
