Question

11.

Answer 1

$\\ \tt\leadsto tan\theta=\dfrac{Perpendicular}{Base}$

$\\ \tt\leadsto tan29=\dfrac{h}{400}$

$\\ \tt\leadsto h=400tan29$

$\\ \tt\leadsto h=221.7ft$

Answer 2

Answer:

221.7 ft (nearest tenth).

Step-by-step explanation:

The situation can be modeled as a right triangle where the base is the distance between the observer (O) and the building (B), and the height is the distance between the base of the building (B) and the kite (K).

To find the height of the kite, use the tan trigonometric ratio.

Tan trigonometric ratio

$\sf \tan(\theta)=\dfrac{O}{A}$

where:

• $\theta$ is the angle
• O is the side opposite the angle
• A is the side adjacent the angle

Given values:

• $\theta$ = 29°
• O = h
• A = OB = 400 ft

Substitute the given values into the formula and solve for h:

$\implies \sf \tan(29^{\circ})=\dfrac{h}{400}$

$\implies \sf h=400\tan(29^{\circ})$

$\implies \sf h=221.7236206...$

Therefore, the height of the kite flying over the building is 221.7 ft (nearest tenth).

Learn more about angles of elevation here:

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