Question

Two Flights Two aeroplanes leave an airport, one after the other. After moving on runway, one flies due North and other flies due South. The speed of two aeroplanes is 400 km/hr and 500 km/hr respectively. Considering PQ as runway and A and B are any two points in the path followed by two planes. Based on this data choose the correct answer: 1) tanθ; if ∠APQ = θ * (A)1/2 (B) - 1/2 (C) 3/2 (D) 3/4 1
2. cot B= * (A) 3/4 (B) 1/5 (C) 3/8 (D) 15/8 3. tan A= * (A) 2 (B) - 2 (C) 4/3 (D) 2/3​

Answer 1

Answer:

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Answer 2

Given:

Speed of first aeroplane = 400km/hr

Speed of second aeroplane = 500km/hr

To find:

1. Tan θ

2. Cot B

3. Tan A

Solution:

From the figure attached below, we can say that ΔAQP is a right-angle triangle.

The length of sides AQ and PQ are given as 1.2km and 1.6km respectively. Then by Pythagoras theorem, we can find the length of hypotenuse AP.

According to Pythagoras theorem,

AP² = PQ² + AQ²

or AP² = 1.6² + 1.2²

or AP = $\sqrt{2.56+1.44}$

= 2

1) In ΔAQP, if ∠APQ = θ,

Tan θ = AQ / PQ = 1.2 / 1.6

= 3 / 4

2) BQP is also a right-angle triangle with lengths of BQ and PQ given as 3km and 1.6km respectively. Using Pythagoras theorem in ΔBQP,

PB² = QB² + PQ²

or PB² = 3² + 1.6²

or PB = $\sqrt{9 + 2.56}$

3.4

Cot B = BQ / PQ = 3 / 1.6

= 30 / 16

= 15 / 8

3) Tan A = PQ / AQ = 1.6 / 1.2

= 8 / 6

= 4 / 3

Hence, tanθ =(D) 3/4 , cotB = (D) 15/8, and  tanA = (C) 4/3