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The angle of elevation of the top \mathrm{P} of a tower from the feet of one person standing due South of the tower is 45^{\circ} and from the feet of another person standing due west of the tower is 30^{\circ}. If the height of the tower is 5 meters, then the distance ( in meters) between the two persons is equal to

Option: 1

10


Option: 2

5 \sqrt{5}


Option: 3

\frac{5}{2} \sqrt{5}


Option: 4

5


Answers (1)

best_answer


Tower \mathrm{AB}=5 \mathrm{~m}
\angle \mathrm{APB}=45^{\circ}
\angle \mathrm{PAB}=90^{\circ}


\tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{AP}}
1=\frac{\mathrm{AB}}{\mathrm{AP}}
\mathrm{AP}=5 \mathrm{~m}

\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AQ}}
\frac{1}{1 \sqrt{3}}=\frac{5}{\mathrm{AQ}}
\mathrm{AQ}=5 \sqrt{3}


\mathrm{AP}^{2}+\mathrm{AQ}^{2}=\mathrm{PQ}^{2}
\mathrm{PQ}^{2}=5^{2}+(5 \sqrt{3})^{2}
\mathrm{PQ}^{2}=25+75=100
\mathrm{PQ}=10 \mathrm{~cm}

Option (A) 10 cm correct.



 

Posted by

SANGALDEEP SINGH

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