Question

General Sherman, a tree located in Sequoia National Park, stands 275 feet tall. To see the top of the tree, Carlos looks up at a 11° angle of elevation. If Carlos is 6 feet tall, how far is he from the base of the tree to the nearest foot?

Answer 1

The distance from the tree to Carlos is 1384 ft.

Step-by-step explanation:

See the attached diagram.

AB = Height of the tree = 275 ft.

CD = Height of Carlos = 6 ft.

So, AE = CD = 6 feet and EB = (AB - AE) = 275 - 6 = 269 ft.

Now, ∠ EDB = 11° {Given}

Taking the right triangle Δ EDB,

$\tan 11^{\circ} = \frac{EB}{ED} = \frac{269}{ED}$

⇒ ED = 1383.88 ft ≈ 1384 ft.

Therefore, the distance from the tree to Carlos is 1384 ft. (Answer)

Answer 2

Answer:

Carlos is 1384 ft away from the base of the tree.

Step-by-step:

Draw a triangle with a height of 269 ft because Carlos is 6 ft tall, so you would do 275- 6= 269. Since you are finding the base you would use tangent. So tan 11 degrees = 269/x, multiply x on both sides to get xtan11=269, then divide both sides by tan 11, and you would get 1383.88503, which rounded to the nearest foot would be 1384.