A satellite flying at height h is watching the top of the two tallest mountains in uttarakhand and karnataka ,them being nanda devi(height 7,816m) and mullayanagiri (height 1,930 m). the angles of depression from the satellite , to the top of nanda devi and mullayanagiri are 30° and 60° respectively. if the distance between the peaks of two mountains is 1937 km , and the satellite is vertically above the midpoint of the distance between the two mountains. Find the distance between peaks and satellite
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In right ∆FGA we have,
→ ∠ F AG = 30°
→ AG = DI = (1937/2) km.
So,
→ cos 30° = AG / AF
→ (√3/2) = (1937/2) / AF
→ √3AF = (1937/2) * 2
→ AF = (1937/1.73) = 1139.4 km (a) (Ans.1)
also,
→ tan 30° = FG/AG
→ (1/√3) = FG/(1937/2)
→ √3FG = (1937/2)
→ FG = (1937√3/6)
→ FG = 569.7 km ------------------- Eqn.(1)
now, In right ∆FHP we have,
→ ∠FPH = 60°
→ HP = IS = (1937/2) km.
So,
→ cos 60° = HP / FP
→ (1/2) = (1937/2) / FP
→ FP = (1937/2) * 2
→ FP = 1937 km (c) (Ans.2)
now,
→ FI = FG + GI
→ FI = FG + AD
putting value from Eqn.(1),
→ FI = 569.7 + 7816
→ FI = 569.7 km + 7816m
→ FI = 569.7 + 7.816
→ FI = 577.52 km (b) (Ans.3)
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