Question

A and B are two points 30 m apart in a line on the horizontal plane through the foot of a tower lying on opposite sides of the tower. The distances of the top of the tower from A and B are 20 m and 15 m respectively, if the angle of elevation of the top of the tower at A is $\theta$ and the height of the tower is h, then

• A. $cos\theta =29/36$
• B. $cos\theta =43/48$
• C. $h=5\sqrt{5}/\sqrt{3}$
• D. $h=5\sqrt{455}/12$

Answer 1

Answer: (B), (D)

The correct options are
B $cos\theta =43/48$
D $h=5\sqrt{455}/12$

Let OP be the tower of height h, then $AB=30,AP=20$ and $BP=15$
so from $\Delta$ABP
$\cos \theta =\frac{{20}^2+{30}^2-{15}^2}{2\times 20\times 30}$
$=\frac{1075}{1200}=\frac{43}{48}$
$\Rightarrow \sin \theta =\sqrt{1-{\left(\frac{43}{48}\right)}^2}$
$=\frac{\sqrt{2304-1849}}{48}$ $=\frac{\sqrt{455}}{48}$
and $h=20\sin \theta =20\times \frac{\sqrt{455}}{48}=\frac{5\sqrt{455}}{12}$