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Question

# Two pillars of equal heights stand on either side of a road which is 100 m wide. At a point on the road between the pillars, the angles of elevation of the tops of the pillars are 60∘ and 30∘. Find the height of each pillar and position of the point on the road. [Take √3=1.732] [3 MARKS]

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Solution

## Let AB and CD be two pillars, each of height h metres and let AC be the road such that AC = 100 m. Let O be the point of observation on AC. Let OA = x metres and OC = (100 - x) m Also, ∠AOB=60∘ and ∠COD=30∘ AB⊥AC and CD⊥AC From right ΔOAB, we have [12 MARK] ABOA=tan60∘=√3 ⇒hx=√3⇒h=√3x In Δ ADE [12 MARK] tan60=hx x√3=h x=h√3……(1) [1 MARK] From right ΔOCD, we have CDOC=tan30∘=1√3 ⇒h(100−x)=1√3⇒h=(100−x)√3 .....(ii) √3x=(100−x)√3⇒3x=(100−x) ⇒4x=100⇒x=25 Putting x = 25 m in (i), we get h=(25×√3)=(25×1.732)=43.3 In Δ BCE tan 30=h100−x 1√3=h100−h√3 from (1) [1 MARK] 1√3=√3h100√3−h 100√3−h=3h 100√3=4h h=25(1.732) =43.3 m [12 MARK] Hence, the height of each pillar is 43.3 m and the point of observation is 25 m away from the first pillar.

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