Trigonometry (9) help~ quick

VINZ

2009-11-26 6:29 am

1. There are 3 buildings, A,B and C, along the side of a lake. The distance A and B that between A and C are both 800m. It is given that the true bearing of B from A is 330 and that of C from A is 120, find the distance beteen B and C.
ANS : 1545m

2. A person on the top of a tower of 30m high looks at a car at an angle of depression of 75. After half a minutes, the person looks at the car again at an angle of depression of 20. Given that the moves in only one direction, how far did the car move in that half a minutes?

3. Prove 3cos^2(90-θ)+1 = 4- 3sin^3(90-θ)

4. Prove tan^2θ-1/tan^2θ+1 = 2sin^2θ-1

回答 (1)

用戶9112

2009-11-26 7:15 am

✔ 最佳答案

(1) The E-W distance between A and B = 800sin(360-330) = 400
The N-S distance between A and B = 800cos(360-330) = 400√3
The E-W distance between A and C = 800cos(120 - 90) = 400√3
The N-S distance between A and C = 800sin(120-90) = 400
The E-W distance between B and C = 400 + 400√3 = 1092.8
The N-S distance between B and C = 400 + 400√3 = 1092.8
BC^2 = 1092.8^2 + 1092.8^2
BC = 1545m
(2) The original distance of the car from the tower = 30/tan75 = 8.038m
The distance of the car from the tower after half a minute = 30/tan20 = 82.424
The distance travelled by the car = 82.424 - 8.038 = 74.39 m
(3) LHS = 3cos^2(90 - θ) + 1
= 3[1 - sin^2(90 - θ)] + 1
= 3 - 3sin^2(90 - θ) + 1
= 4 - 3sin^2(90 - θ) = RHS
(4) LHS = (tan^2θ - 1)/(tan^2θ + 1)
= (sin^2θ/cos^2θ - 1)/(sin^2θ/cos^2θ + 1)
= (sin^2θ - cos^2θ)/(sin^2θ + cos^2θ)
= sin^2θ - cos^2θ
= sin^2θ - (1 - sin^2θ)
= 2sin^2θ - 1 = RHS

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