Question

.) एक हवाई जहाज का पायलेट पूर्व दिशा में 100 किमी. दूर स्थित लक्ष्य पर 20 मिनट में पहुंचना चाहता है। यदि वायु उत्तर-पूर्व की ओर से 15 किमी./घण्टा के वेग से आ रही है तो वायु के सापेक्ष हवाई जहाज का वेग ज्ञात 15 करो।​

Answer 1

Answer:

The line W ← → E is the desired course flown. AB represents the plane’s velocity, at an unknown angle θ south of east, and BC is the wind’s velocity. The addition of these vectors gives the resultant AC which is the velocity made good.

So, now, all we need to do is solve for θ. In dotted lines I have shown the line BD which is perpendicular to the W - E line.

In the ΔBCD, BD/BC = sin45° → BD = BC.sin45° = 100.sin45°.

In ΔABD, sinθ = BD/AB → θ = arcsin(100.sin45°/450)

θ = 9.04°, so the plane needs to fly 9° South of East (099°T).

The length AC is the speed at which the plane is travelling over the ground. You can use the same two triangles to find AC. (Hint, find AD using cosθ, then subtract CD using cos45°)