A satellite flying at a 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 (DS) between the two mountains is 1937 km, and the satellite is vertically above the midpoint of the distance between the two mountains, then answer the following questions:
(1) Find the distance of the satellite from the top of Mullayanagiri?
(ii) Find the ratio of distances of the satellite from the top of Mullayanagiri to the top of Nanda Devi?
(1) Find the distance of the satellite from the top of Mullayanagiri?
(ii) Find the ratio of distances of the satellite from the top of Mullayanagiri to the top of Nanda Devi?
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In right AFGA we have,
FAG = 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 AFHP we have,
→ ZFPH = 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),
→ FI569.7 + 7816
→ FI= 569.7 km + 7816m
→ FI = 569.7 + 7.816
→ FI = 577.52 km (b) (Ans.3)