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Mathematics

1 answer:

andre [41]2 years ago

8 0

Answer:

○ $\displaystyle sin\:θ = \frac{12}{13}$

Explanation:

This should help you figure out why this is the answer:

$\displaystyle \frac{OPPOSITE}{HYPOTENUSE} = sin\:θ \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:θ \\ \frac{OPPOSITE}{ADJACENT} = tan\:θ \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:θ \\ \frac{HYPOTENUSE}{OPPOSITE} = csc\:θ \\ \frac{ADJACENT}{OPPOSITE} = cot\:θ$

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A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place

Kazeer [188]

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = $\frac{AB}{BC} = \frac{h}{x}$

Taking inverse trigonometric function, α = tan⁻¹ ( $\frac{h}{x}$ ) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

$\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}$

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and $\frac{dh}{dt}$ = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

$\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512$