Question

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, then answer the following questions.

(i) Find tanθ; if ∠APQ = θ.
(ii) Find cotB.
(iii) Find tanA.
(iv) Find secA.
(v) Find cosecA.


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Answer 1

《1》tanθ=opposite/adjacent = 1.2/1.6 = 3/4

《2》cotB=adjacent/opposite = 3/1.6 = 15/8

《3》tanA=opposite/adjacent=1.6/1.2=4/3

《4》AP²=AQ²+QP²=(1.2)²+(1.6)²

=1.44+2.56=4

AP=root4=2

SecA=Hypotunues/adjacent = 2/1.2 = 5/3

《5》BP²=BQ²+QP²=3²+1.6²=9+2.56

=11.56

BP=root 11.56 = 3.4

CosecB=hypotenuse/opposite=3.4/1.6=17/8

Answer 2

Given:

Speed of first aeroplane = 400km/hr

Speed of second aeroplane = 500km/hr

To find:

The values of Tan θ if ∠APQ = θ, Cot B, Tan A, Sec A, Cosec A.

Solution:

From the figure, it can be seen 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,

$AP^{2}=PQ^{2} +AQ^{2}$

$AP^{2}=1.6^{2} +1.2^{2}$

$AP^{2}=2.56+1.44$

$AP^{2} =4$

Taking square root on both sides,

$AP=2km$

In ΔAQP, if ∠APQ = θ,

$Tan\theta=\frac{OppositeSide}{AdjacentSide}$

$Tan\theta=\frac{AQ}{PQ}$

$Tan\theta=\frac{1.2}{1.6}$

∴ $Tan\theta=0.75$

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

$PB^{2} =QB^{2} +PQ^{2}$

$PB^{2} =3^{2} + 1.6^{2}$

$PB^{2} =9+2.56$

$PB^{2}=11.56$

Taking square root on both sides,

$PB=3.4km$

The Cotangent of an angle is the reciprocal of the tangent value of the same angle. Here, ∠B is the required angle.

$CotB=\frac{1}{TanB}$

$CotB=\frac{AdjacentSide}{OppositeSide}$

$CotB=\frac{QB}{PQ}$

$CotB=\frac{3}{1.6}$

$CotB=1.875$

Now, we need to find the remaining trigonometric ratios using ∠A.

$TanA=\frac{OppositeSide}{AdjacentSide}$

$TanA=\frac{PQ}{AQ}$

$TanA=\frac{1.6}{1.2}$

$TanA=1.33$

The secant of an angle is the reciprocal of the cosine of the same angle.

$SecA=\frac{1}{CosA}$

$SecA=\frac{Hypotenuse}{AdjacentSide}$

$SecA=\frac{AP}{AQ}$

$SecA=\frac{2}{1.2}$

$SecA=1.66$

The cosecant of an angle is the reciprocal of the sine of the same angle.

$CosecA=\frac{Hypotenuse}{OppositeSide}$

$CosecA=\frac{AP}{PQ}$

$CosecA=\frac{2}{1.6}$

$CosecA=1.25$

The following are the answers:

(i) tanθ = 0.75, if ∠APQ = θ.

(ii) cotB = 1.875

(iii) tanA = 1.33

(iv) secA = 1.66

(v) cosecA = 1.25