SAT Math · Right triangles and trigonometry · medium

A person standing 50 feet from the base of a tower observes the top of the tower at an angle of elevation of 40°. To the nearest tenth of…

40° 50 ft height
A person standing 50 feet from the base of a tower observes the top of the tower at an angle of elevation of 40°40°. To the nearest tenth of a foot, what is the height of the tower? (Use tan⁡(40°)≈0.839\tan(40°) \approx 0.839.)

Answer and explanation

Answer: 42.0 or 41.9 or 41.95

The tower, the ground, and the line of sight form a right triangle. The angle of elevation is 40°40°, the adjacent side is 50 feet, and the opposite side is the height. Tangent connects them: tan⁡(40°)=h50\tan(40°) = \frac{h}{50}, so h=50×0.839=41.95h = 50 \times 0.839 = 41.95, which rounds to 42.0 feet. A common wrong answer is 38.3, which comes from using sine or cosine incorrectly (such as 50cos⁡40°≈38.350 \cos 40° \approx 38.3). Another is 59.6, from dividing 500.839\frac{50}{0.839}, which inverts the relationship. Choose the ratio that links opposite and adjacent.

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