Ex 9.1, 4 - The angle of elevation of top of a tower - Questions easy to difficult

EX 9.1, 4 - Part 2

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Ex 9.1 , 4 The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower. Let tower be AB Let point be C Distance of point C from foot of tower = 30m Hence, BC = 30m Angle of elevation = 30° So, ∠ACB = 30° Since tower is vertical, ∠ ABC = 90° We need to find height of tower i.e. AB In right triangle ABC tan C = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" c " )/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶) tan 30° = (" " AB)/BC 1/√3 = (" " AB)/30 30/√3 = AB AB = 30/√3 Multiplying √3 in numerator and denominator AB= 30/√3 × √3/√3 AB = (30√3)/3 AB = 10√3 Hence, Height of tower = AB = 10√3 m

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