Hot air balloon is attached to the ground by a cable that is 200 meters long. The cable makes a 57° angle with the ground. How many meters above the ground is the balloon if there is no slack in the cable? Round your answer to the nearest tenth of a meter

By inscribing the given information in a right triangle, we will see that the ballon is 167.7 meters above the ground.

Think in the situation as in a right triangle, you have a hypotenuse that is equal to the length of the cable, you know one angle of 57°, and the opposite cathetus to this angle represents the height at which the ballon is, this is what we want to find.

Using the relation:

sin(θ) = (opposite cathetus)/hypotenuse.

Where:

Replacing that we get:

sin(57°) = height/200m

sin(57°)*200m = height = 167.7m

So the balloon is 167.7 meters above the ground.

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Let C be the position of the meteorological station, B the position of the balloon and CB be the cable.

Let the height of the balloon from the ground be AB=hmetres.

In a right angled triangle ACB

BC

AB

=sin60

∘

⇒

200

h

=

2

3

since

sin60

∘

=

2

3

⇒h=200×

2

3

⇒h=100

3

∴h=100×1.732=173.2 since

3

=1.732

Hence, the height of the balloon above the ground is 173.2m