Question

The angle of elevation of the top of a pole as seen from a point 13 ft away from the base is double its angle of elevation as seen from a point 47 ft farther from the pole. Find the height of the pole above the level of the observer's eyes. 10) A) 34 ft B) 86 ft C) 60 ft D) 45 ft

Answer 1

The height of the pole above the level of the observer's eyes is 31.42 ft

Let the angle of elevation of the top of the pole from a point 47 ft away be α.

Let the angle of elevation of the top of the pole from the point 13 ft away be β.

Since the angle of elevation of a point 13 feet away is twice that from 47 ft away, we have that β = 2α.

If the height of the pole above the level of the observer's eye is h, we have that

tanβ = tan2α = h/13 and tanα = h/47

From trigonometric identities tan2α = 2tanα/(1 - tan²α)

Substituting the values of the variables into the equation, we have

tan2α = 2tanα/(1 - tan²α)

h/13 = 2h/47 ÷ [1 - (h/47)²]

Dividing through by h, we have

1/13 = 2/47 ÷ [1 - (h/47)²]

Cross-multiplying, we have

1/13 = 2/47 ÷ [1 - (h/47)²]

1 - (h/47)² = 2/47 × 13

1 - (h/47)² = 26/47

(h/47)² = 1 - 26/47

(h/47)² = (47 - 26)/47

(h/47)² = 21/47

Taking square-root of both sides, we have

h/47 = √(21/47)

Cross-multiplyng, we have

h = √(21/47) × 47

h = √(21 × 47)

h = √987

h = 31.42 ft

The height of the pole above the level of the observer's eyes is 31.42 ft

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