Question

Sara is flying a kite at the park. The angle of elevation from Sara to kite in the sky is 53° and the length of the string is 32 feet. Of Sara is 5 feet tall how far off the ground is the kite

Answer 1

Answer:

Using sine ratio:

$\sin \theta = \frac{\text{Opposite side}}{\text{Hypotenusesside}}$

As per the statement:

The angle of elevation from Sara to kite in the sky is 53° and the length of the string is 32 feet. and Sara is 5 feet tall.

⇒Angle of elevation( $\theta$) = 53° and

Hypotenuse side = 32 feet (As you can see the figure as shown below)

then;

Apply the sine ratio in triangle ACE we have;

$\sin 53^{\circ} = \frac{CE}{32}$

$0.79863551004 = \frac{CE}{32}$

Multiply both sides by 32 we get;

$0.79863551004 \cdot 32 = CE$

Simplify:

CE = 25.5563363215 feet.

Then:

DE = CD+CE = 5 + 25.5563363215 = 30.5563363215 ≈30.6 feet.

Therefore,  30.6 feet far off the ground is the kite.

Answer 2

Answer: The answer is 30.56 feet.

Step-by-step explanation:  As shown in the attached figure, Sara is flying a kite at the park with angle of elevation from Sara to kite in the sky, ∠BAC = 53° and length of the string, AC = 32 feet. Sara is 5 feet tall. We need to find the height of the kite from the ground.

Fro the right angled-triangle ABC, we have

$\sin \angle BCA=\dfrac{AB}{AC}\\\\\\\Rightarrow \sin 53^\circ=\dfrac{p}{32}\\\\\\\Rightarrow p=32\times \sin 53^\circ\\\\\Rightarrow p=25.56.$

Since Sara is 5 feet tall, so the height of the kite from the ground is

AD = AB + BD = p + 25.56 = 5 + 25.56 = 30.56 feet.

Thus, the answer is 30.56 feet.