# Math: Trigonometry Word Problem!?

A rescue helicopter flew from its home base for 35km on a course of 330 degrees to pick up an accident victim. It then flew 25km on a course of 090 degrees to the hospital. What distance and on what course will the helicopter fly to return directly to its home base?

### 2 Answers

- MarkLv 61 decade agoFavorite Answer
Let the home base be B, the accident victim be A and the hospital H.

We are told AB = 35 km, AH = 25 km and ∠BAH = 60°

Using the cosine rule in triangle BAH

HB² = 35² + 25² - 2 * 35 * 25 * cos 60°

HB² = 975

HB = 31.2 km

Using the sine rule in triangle BAH

HB / sin 60° = 25 / sin θ where θ = angle ABH

sin θ = 25 * sin 60° / 31.2 = 0.6934

θ = 43.9°

Then θ - 30° = 13.9°

Hence the helicopter's course should be on a bearing of 180 + (θ - 30°) = 193.9°

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- PamelaLv 44 years ago
Didn't check my mail earlier, just saw your message =D Anyways, I still stick by my original answer. The reason was this: The altitude Y of the kite could be calculated the way you suggested, but it'd be a lot more difficult. The simpler way would be to first assume ground level to be where Dennis is holding the kite, and then when you find the kite's altitude from that ground level, add the altitude of where Dennis is holding the kite. Basically, you split Y into two parts: the triangle's leg & an altitude identical to the one that Dennis is holding the kite at. Sorta hard to explain... I'll give it again anyways =D; Let a be the altitude of where Dennis is holding the kite. X = 50 ÷ cos30 = 50 ÷ (√3 ÷ 2) = (100√3) ÷ 3 Y = (50 × tan30) + a = 50 × (1 ÷ √3) + a = (50√3) ÷ 3 + a

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