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An Aeroplane When 3,000 Meters High Passes Vertically Above Another Aeroplane at an Instance When Their Angles of Elevation at the Some Observation Point Are 60° and 45° Respectively. - Mathematics

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प्रश्न

An aeroplane when 3,000 meters high passes vertically above another aeroplane at an instance when their angles of elevation at the same observation point are 60° and 45° respectively. How many meters higher is the one than the other?

योग

उत्तर

Let P1 and P2 denote the positions of the two planes. Then in right-angled ΔP1AB, 

`(P_1B)/(AB) = tan 45° ⇒ P_1B = AB`  

In right-angled ΔP2AB, 

⇒ `(P_2B)/AB = tan 60° = sqrt3`

⇒ AB = `(P_2B)/sqrt3`

⇒ AB = `3000/sqrt3`

⇒ AB = 1000√3

∴ Vertical distance between the two planes is P1P2 = P2B - P1B

= 3000 - 1000√3
= 1000( 3 - √3) m

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Heights and Distances - Solving 2-D Problems Involving Angles of Elevation and Depression Using Trigonometric Tables
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Trigonometry - Exercise 4

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आईसीएसई Mathematics [English] Class 10
अध्याय 18 Trigonometry
Exercise 4 | Q 10

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