The horizontal distance between two poles is 15 m. The angle of depressionof the top of first pole as seen from the top of second pole is 30o. If theheight of the second pole is 24 m, find the height of the first pole. $|Us e \sqrt{3} = 1.732|$

Solution

$Let Ab and CD be the two poles, where CD (the second pole) = 24 m . BD = 15 m Let the height of pole AB be h m Al = BD = 15 m and AB = LD = h So, CL = CD - LD = 24 - h In ∆ ACL, \tan 30 ° = \frac{CL}{AL} \Rightarrow \tan 30 ° = \frac{24 - h}{15} \Rightarrow \frac{1}{\sqrt{3}} = \frac{24 - h}{15} \Rightarrow 24 - h = \frac{15}{\sqrt{3}} = 5 \sqrt{3} \Rightarrow h = 24 - 5 \sqrt{3} \Rightarrow h = 24 - 5 x 1.732 (Taking \sqrt{3} = 1.732) \Rightarrow h = 15.34 Thus, height of the first pole is 15.34 m .$

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Some Applications Of Trigonometry

The horizontal distance between two poles is 15 m. The angle of depressionof the top of first pole as seen from the top of second pole is 30o. If theheight of the second pole is 24 m, find the height of the first pole. $|Us e \sqrt{3} = 1.732|$

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Some Applications Of Trigonometry

The horizontal distance between two poles is 15 m. The angle of depressionof the top of first pole as seen from the top of second pole is 30o. If theheight of the second pole is 24 m, find the height of the first pole. Use 3 = 1.732

Some More Questions From Some Applications of Trigonometry Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The horizontal distance between two poles is 15 m. The angle of depressionof the top of first pole as seen from the top of second pole is 30o. If theheight of the second pole is 24 m, find the height of the first pole. $|Us e \sqrt{3} = 1.732|$