Question

SOLVE THIS ANYONE
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Answer 1

Explanation:

Height of the jet plane = 1500√3 m

Initially the jet plane is at A and after 15 seconds flight at B

ED = x , EC = y , where D is right below A and C is right below B .

( The position of the plane after 15 sec )

From ∆EBC, Cot 30° = y / 1500√3

= y/ 1500√3 = √3

= y = 1500√3 × √3 = 4500m

From ∆ EDA , Cot 60° = x / 1500√3

x / 1500√3 = 1 / √3

= x = 1500m

AB = DC = Distance travelled in 15 sec

= y - x = ( 4500- 1500) m

= 3000m

Distance travelled in 1 sec

= 3000/ 15 m

= 200m

Hence the speed of the jet plane = 200

m/ sec = 200/ 1000 × 60 × 60 km/hr

= 720km/ hr

Hope it helps you

Answer 2

Answer:

$hope \: it \: helps \:$