A toy rocket is launched from a platform 33 feet above the ground at a speed of 83 feet per second. The height of the rocket in
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6% = 1200
9% = 2400
12% = 3800
6% = 1200
9% = 2400
12% = 3800
1)
∠BAC = ∠NAC - ∠NAB = 144 - 68 = 76⁰
AB = 370 m
AC = 510 m
To find BC we can use cosine law.
a² = b² + c² -2bc*cos A
|BC|² = |AC|²+|AB|² - 2|AC|*|AB|*cos(∠BAC)
|BC|² = 510²+370² - 2*510*370*cos(∠76⁰) =
|BC| ≈ 553 m
2)
To find ∠ACB, we are going to use law of sine.
sin(∠BAC)/|BC| = sin(∠ACB)/|AB|
sin(76⁰)/553 m = sin(∠ACB)/370 m
sin(∠ACB)=(370*sin(76⁰))/553 =0.6492
∠ACB = 40.48⁰≈ 40⁰
3)
∠BAC = 76⁰
∠ACB = 40⁰
∠CBA = 180-(76+40) = 64⁰
Bearing C from B =360⁰- 64⁰-(180-68) = 184⁰
4)
Shortest distance from A to BC is height (h) from A to BC.
We know that area of the triangle
A= (1/2)|AB|*|AC|* sin(∠BAC) =(1/2)*370*510*sin(76⁰).
Also, area the same triangle
A= (1/2)|BC|*h = (1/2)*553*h.
So, we can write
(1/2)*370*510*sin(76⁰) =(1/2)*553*h
370*510*sin(76⁰) =553*h
h= 370*510*sin(76⁰) / 553= 331 m
h=331 m
