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Find the angle of elevation of a point on the ground to the top of a tower if the horizontal distance between them is √3 times the height of tower. |
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Answer» =34.60m Step-by-step explanation: To find → Height of tower (AB) Let AB =h m In △ABC, by Trigonometry tan60 ∘ = AB
3 = BC h ⇒BC=( 3
h )m Since BCDE is a ||gm, so BC=DE and CD = BE ∴DE=( 3
h )m;BE=40m Now in △ADE tan30 ∘ = DE AE
3
1 = 3
h
h−40
h=3h−120 2h=120m h=60m Height of tower = 60 m. To find → HORIZONTAL Distance from point of observation (BC) BC= 3
h = 3
60 =20 3 m BC=20 3 m=34.60m |
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