Question

II. Two jet planes X and Y leave an airport, one after the other. After moving on runway, one flies to North and other flies to South. The speed of aeroplanes X and Y is 400 km/hr and 508 km/hr respectively. Considering PQ as runway and A and B are any two points in the path followed by two planes, answer the following questions: 1.5km! 2.5km 3km

the value of sin B in ∆PQB is
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Answer 1

Step-by-step explanation:

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Answer 2

Given:

The speed of north flying airplane (X) = 400km/h

The speed of south flying airplane (Y) = 508km/h

To Find:

Distance between the planes after 1.5 hours (BC)

Solution:

The formula to calculate speed is,

         Speed = $\frac{Distance}{Time}$

          Distance = Speed × Time

   Now, ∠BAC = 90°   ( North and South are perpendicular to each other)

By Pythagoras theorem in ΔABC,

 (BC)² = (AB)²+(AC)²

 (BC)² = (400)²+(508)²

 (BC)² = 160000 + 258064

    BC = √418064

    BC = 646.57 km.

Therefore, two jet planes would be at a distance of 646.57km.