Question

Question 28.... plz answer fast

Answer 1

Answer:

Height of tower = 70.95

Distance = 70.95

Answer 2

In the adjoining figure,
AC is a tower, from a point B on the ground, the angle of elevation of AC is 45°. On moving 30m towards the tower, the angle of elevation of the top of it from point D on the ground is 60°.

In triangle ABC,

$\tan(45°) = \frac{AC}{BC}$

$1 = \frac{AC}{BC}$

$AC = BC$ (1)

In triangle ADC,

$\tan(60°) = \frac{AC}{DC}$

$\sqrt{3} = \frac{AC}{DC}$

$DC\sqrt{3} = AC$ (2)

From (1) and (2)

$BC = DC\sqrt{3}$
$BD + DC = DC\sqrt{3}$
$BD = DC(\sqrt{3} -1)$
$30 = DC(\sqrt{3} -1)$
$15(\sqrt{3} +1) = DC$

Approximately, DC = 41m.

Therefore BC = 71m
Original distance from foot of the tower = 71m.

As AC = BC = Height of tower.

Height of tower = 71m