Question

IZ) Samantha wants to determine the height of a flagpole at school. Her eye level is 4.6 feet from the ground and
she stands 26 feet from the flagpole. If the angle of elevation is about 68°, what is the height of the flagpole to
the nearest tenth foot?

Answer 1

Answer:

69 feet.

Step-by-step explanation:

See the attached diagram.

AB is the height of the flagpole and DE is the height of Samantha.  

Now, ∠ CEB = 68°

Now, AC = DE = 4.6 feet, and CE = AD = 26 feet {Given}

Then, $\tan 68 = \frac{CB}{CE} = \frac{CB}{26}$

⇒ CB = 26 tan 68 = 64.35 feet.

Now, height of the flagpole is AB = AC + CB = 4.6 + 64.35 = 68.95 feet ≈ 69 feet. (Answer) (Approximate}