Question

The height of the building is feet long.

• A. 55+(50/$\sqrt{3}$)
• B. 55feet
• C. 10$\sqrt{3}$ feet

Answer 1

The correct option is A 55+(50/$\sqrt{3}$)

The height of the adjacent building is = AB +BC + CD

We know that BC = height of Sam = 5 feet

CD = height of Sam’s building = 50 feet

Consider the right triangle ∆ABO,

∠ABO = 90°

It is given that, OB = Distance between the buildings = 50 feet

Therefore, $\frac{AB}{OB}=tan30°$

AB = OB$\times tan{30}^{\circ }$

= $\frac{50}{\sqrt{3}}$ feet

Therefore the height of the adjacent building = $\frac{50}{\sqrt{3}}$ + 50 + 5 = (55 + $\frac{50}{\sqrt{3}}$)

= (55 + $\frac{50\sqrt{3}}{3}$) feet

= (55 + $\frac{50}{\sqrt{3}}$) feet