Question

Max observes the zoo and the library from a helicopter flying at a height of 200 times square root of 3 feet above the ground, as shown below: A helicopter is flying at a height of 200 multiplied by square root of 3 feet above the ground. A zoo and a library are on the ground on the same side of the helicopter. The angle made by the line joining the helicopter and the zoo with the ground is 60 degrees. The angle made by the line joining the helicopter and the library with the ground is 30 degrees. What is the distance between the zoo and the library? (1 point) HELPPPP!!!
400 feet

200 feet

600 feet

800 feet

Answer 1

Answer:

The distance between the zoo and the library is 400ft

Step-by-step explanation:

Let x be the distance between point G and the zoo

Let y be the distance between the zoo and the library

To solve this problem quickly, we use trigonometric relations

tan(angle) = opposite cathetus / adjacent cathetus

In this case

tan(30) = (200√3) / (x+y)

(x+y) =  (200√3)/ tan(30)

(x+y) = 600 ft

For the other angle

tan(60) = (200√3) / (x)

(x) =  (200√3)/ tan(60)

x = 200 ft

y = 600 ft - 200 ft = 400 ft

The distance between the zoo and the library is 400ft

Answer 2

Answer:

The correct option is A.

Step-by-step explanation:

Let the distance between the zoo and the library is x.

In a right angled triangle,

$\tan \theta = \frac{opposite}{adjacent}$

In triangle ABG,

$\tan (30^{\circ}) = \frac{AG}{GB}$

$\frac{1}{\sqrt{3}} = \frac{200\sqrt{3}}{GB}$

On cross multiplication, we get

$GB\times 1=\sqrt{3}\times 200\sqrt{3}$

$GB=600$

In triangle AGC,

$\tan (60^{\circ}) = \frac{AG}{GC}$

$\sqrt{3} = \frac{200\sqrt{3}}{GC}$

On cross multiplication, we get

$GC\times \sqrt{3}=200\sqrt{3}$

Divide both sides by √3.

$GC=200$

The distance between the zoo and the library is

$CB=GB-GC$

$x=600-200$

$x=400$

The distance between the zoo and the library is 400 feet. Therefore the correct option is A.