# Question deeb3

Oct 14, 2016

In figure
AB = h ft = Height of the building
C is the point on the ground from which angle of elevation of building $\angle A C B = {30}^{\circ} 20 ' = {\left(30 + \frac{20}{60}\right)}^{\circ} = {\left(\frac{91}{3}\right)}^{\circ}$

D is the point on the ground(50 ft closer to the building from C) from which angle of elevation of building $\angle A D B = {45}^{\circ}$
Let BD = x ft

Now $\frac{A B}{B D} = \frac{h}{x} = \tan \angle A D B = \tan {45}^{\circ} = 1$

$\implies \frac{h}{x} = 1 \implies x = h$

Again$\frac{A B}{B C} = \frac{A B}{B D + D C} = \frac{h}{x + 50} = \tan \angle A C B = \tan {\left(\frac{91}{3}\right)}^{\circ}$

$\implies \frac{h}{x + 50} = 0.585$

$\implies \frac{h}{h + 50} = 0.585$

$\implies h = \left(h + 50\right) \times 0.585$

$\implies h \left(1 - 0.585\right) = 50 \times 0.585 \approx 29.25$

$h = \frac{29.25}{0.415} \approx 70.5 f t$

Feb 5, 2017

Height of building$= 70.5 f t$

#### Explanation:

Let line AB be the building with A the top.
Let C be the first point with ascending angle 30°20'
Let D be the second point with ascending angle 45°

In triangle ACD angle D = 180°-45°=135°

:.180°-(135°+30°20')=14°40'= angle A

In triangle ACD:

(AD)/(Sin30°20')=50/(sin14°40')

Multiply both sides by sin30°20'

:.cancel(sin30°20')/1 xx ( AD)/cancel(sin30°20')=(50 xx sin30°20')/(sin14°40')

$\therefore A D = \frac{50 \times 0.505029841}{0.253195168}$

$\therefore A D = \frac{25.25149208}{0.253195168}$

$\therefore A D = 99.73133484 = h y p o t e \nu s e$

In triangle ADB angle D = 45°

AB=height of building=opposite side

:.sin45°=(opposite)/(hypotenuse)#

$\therefore A B = \sin 45 \times 99.73133484$

$\therefore A B = 70.5 f t$