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  • The angle of elevation of the top of a vertical tower PQ of height 10 from a point <img alt="\mathrm{A}" src="/latex-image/?%5Cmathrm%7

The angle of elevation of the top \mathrm{P} of a vertical tower PQ of height 10 from a point \mathrm{A} on the horizontal ground is 45^{\circ}. Let \mathrm{R} be a point on \mathrm{A Q} and from a point \mathrm{B}, vertically above\mathrm{ R}, the angle of elevation of \mathrm{\mathrm{P} \: is\: 60^{\circ}}. If \mathrm{\angle \mathrm{BAQ}=30^{\circ}, \mathrm{AB}=\mathrm{d}} and the area of the trapezium \mathrm{PQRB}$ is $\alpha, then the ordered pair (\mathrm{d}, \alpha) is :
 

Option: 1

(10(\sqrt{3}-1), 25)
 


Option: 2

\left(10(\sqrt{3}-1), \frac{25}{2}\right)

 


Option: 3

(10(\sqrt{3}+1), 25)
 


Option: 4

\left(10(\sqrt{3}+1), \frac{25}{2}\right)


Answers (1)

\begin{aligned} &\frac{P Q}{A Q}=\tan 45^{\circ}\\ &\Rightarrow A Q=P Q=10\\ &\text { Now } A R+R Q=A Q\\ &d \cos 30^{\circ}+x=10 \ldots \text { i }\\ &\text { Also, } \frac{10-d \sin 30^{\circ}}{x}=\tan 60^{\circ}\\ &x=\frac{10-\frac{d}{2}}{\sqrt{3}} \end{aligned}

\begin{aligned} &\text{Put the in (i)}\\ &\mathrm{ \frac{\sqrt{3} d}{2}+\frac{10-\frac{d}{2}}{\sqrt{3}}=10 }\\ \Rightarrow & \mathrm{\frac{3 d+20-d}{2 \sqrt{3}}=10} \\ \Rightarrow & \mathrm{2 d+20=20 \sqrt{3}} \\ \Rightarrow & \mathrm{2 d=20(\sqrt{3}-1)} \\ \Rightarrow & \mathrm{d=10(\sqrt{3}-1)}\\ & \text{Now area of PQRB}\\ &=\mathrm{\frac{1}{2}(d \sin 30+10) \cdot x} \\ &=\mathrm{\frac{1}{2}(5(\sqrt{3}-1)+10) \cdot \frac{1}{\sqrt{3}}(10-5(\sqrt{3-1})) }\\ &=\mathrm{\frac{1}{2}(5 \sqrt{3}+5) \cdot \frac{1}{\sqrt{3}}(15-5 \sqrt{3})} \\ &=\mathrm{\frac{1}{2 \sqrt{3}} \cdot 5(\sqrt{3}+1) \cdot 5 \sqrt{3}(\sqrt{3}-1) }\\ &=\mathrm{\frac{5.5}{2}(3-1)} \\ &=\mathrm{25}\\ &\therefore \text { option (A) } \\ \end{aligned}

 

Posted by

Sumit Saini

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