First slide

Applications of trigonometry

Question

The angles of elevation of a flying kite at three points A,B,C on a horizontal line (lying in the same vertical plane with the kite) are in the ratio 1 : 2 : 3. If AB=a, BC=b, then the height of the kite is

Difficult

A

$\frac{a}{2 b} \sqrt{(a + b) (3 b - a)}$
B

$\frac{a}{b} \sqrt{(a - b) (3 b + a)}$
C

$\frac{2 a}{b} \sqrt{(a - b) (3 b + a)}$
D

$\frac{a}{b} \sqrt{(a + b) (3 b - a)}$

Solution

Let Q=kite, A=first point observation, B=second point observation, C= third point of observation. Given AB =a, BC=b. Let PQ = h. Since exterior angle = sum of the opposite interior angles,
$∠ P B Q = ∠ B Q A + ∠ B A Q$ $a n d ∠ P C Q = ∠ C B Q + ∠ C Q B$
$∴ A B = a = Q B$
$A p p l y i n g \sin e r u l e$ $i n Δ B Q C$ , $w e g e t \frac{B Q}{\sin ∠ Q C B} = \frac{B C}{\sin ∠ C Q B}$
$\Rightarrow \frac{a}{\sin (180^{0} - 3 θ)} = \frac{b}{\sin θ}$
$\Rightarrow 4 \sin^{2} θ = 3 - \frac{a}{b} = \frac{3 b - a}{b} \Rightarrow \sin^{2} θ = \frac{3 b - a}{4 b}$
$∴ \cos^{2} θ = 1 - \sin^{2} θ = 1 - \frac{3 b - a}{4 b} = \frac{4 b - 3 b + a}{4 b} = \frac{b + a}{4 b}$
$F r o m P B Q$ ,
$\sin 2 θ = \frac{h}{a} \Rightarrow 2 \sin θ \cos θ = \frac{h}{a} \Rightarrow 4 \sin^{2} θ \cos^{2} θ = \frac{h^{2}}{a^{2}}$
$\Rightarrow 4. \frac{3 b - a}{4 b} . \frac{a + b}{4 b} = \frac{h^{2}}{a^{2}} \Rightarrow h^{2} = \frac{a^{2}}{4 b^{2}} (3 b - a) (a + b)$
$∴ h = \frac{a}{2 b} \sqrt{(3 b - a) (a + b)}$

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