Question

The angle of elevation of the top of a tower standing on a horizontal plane from the two points lying
on a line passing through the foot of the tower at distances a and b respectively are complementary
angles. if the line joining the two points sutends an angle o at the top of the tower, then sine is​

Answer 1

Answer:

Let tower of heigh h .

Now see diagram

$\tan( \alpha ) = \frac{h}{a} \\ \\ \tan( \alpha + \theta) = \frac{h}{b} \\ divide \: \: both \: \: eliminate \: \: h \\ \\ a \tan( \alpha ) = b \tan( \alpha + \theta)$

Given that 90 - α and 90-α-0 are complimentary angles

$90 - \alpha + 90 - \alpha - \theta = 90 \\ \alpha = \frac{\pi}{4} - \theta$

So

$a \tan( \frac{\pi}{4} - \theta) = b \tan(\frac{\pi}{4} - \theta + \theta) \\ \\ a( \frac{1 + \tan( \theta) }{1 - \tan( \theta) } ) = b \\ \\ \tan( \theta) = \frac{a + b}{b - a} \\ \\ \sin( \theta) = \frac{a + b}{ \sqrt{ {(a + b)}^{2} + {(b - a)}^{2} } } = \frac{a + b}{ \sqrt{2( {a}^{2} + {b}^{2} )} }$

Answer 2

$\sin\theta=\dfrac{b-a}{b+a}$

Step-by-step explanation:

Let the height of tower be h

Please refer attachment for figure.

∠C and ∠D are complementary angle.

Therefore,  ∠C + ∠D = 90°

In ΔABC, ∠B = 90°

$\tan \alpha=\dfrac{BC}{AB}$

$\tan \alpha=\dfrac{a}{h}$ -----------(1)

In ΔABD, ∠B = 90°

$\tan \alpha=\dfrac{BD}{AB}$

$\tan (\alpha+\theta)=\dfrac{b}{h}$ -----------(2)

Eliminate h from eq(1) and eq(2)

$a\tan(\alpha+\theta)=b\tan \alpha$ -------- (3)

$\angle C=90^\circ-\alpha$

$\angle D=90^\circ-(\alpha+\theta)$

But ∠C + ∠D = 90°

$90^\circ-\alpha+90^\circ-(\alpha+\theta)=90^\circ$

$2\alpha=90^\circ-\theta$

$\alpha=45^\circ-\dfrac{\theta}{2}$

Substitute alpha in equation 3

$a\tan(45^\circ-\dfrac{\theta}{2}+\theta)=b\tan (45^\circ-\dfrac{\theta}{2})$

$a\tan(45^\circ+\dfrac{\theta}{2})=b\tan (45^\circ-\dfrac{\theta}{2})$

$a\cdot \dfrac{1+\tan(\frac{\theta}{2})}{1-\tan(\frac{\theta}{2})}=b\cdot \dfrac{1-\tan(\frac{\theta}{2})}{1+\tan(\frac{\theta}{2})}$

$\dfrac{a}{b}=\left(\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}\right)^{2}$

where, $\dfrac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}=\sqrt{\dfrac{1-\sin\theta}{1+\sin\theta}}$

$\dfrac{a}{b}=\dfrac{1-\sin\theta}{1+\sin\theta}$

Using componendo and Dividendendo

$\dfrac{a+b}{a-b}=\dfrac{2}{-2\sin\theta}$

$\sin\theta=\dfrac{b-a}{b+a}$

Learn more:

Trigonometry identity

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