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2 years ago

In ΔRST, the measure of ∠T=90°, RT = 33, SR = 65, and TS = 56. What is the value of the sine of ∠S to the nearest hundredth?

Mathematics

1 answer:

Maurinko [17]2 years ago

3 0

Answer:

Sin (∠S) = 0.51 (to the nearest hundredth)

Step-by-step explanation:

From the diagram, Let ∠S = θ

Therefore: Sin θ = $\frac{Opposite}{Hypothenuse}$

Where:

Opposite = 33

Hypothenuse = 65

∴ Sin θ = $\frac{33}{65}$ = 0.51

∴ Sin θ = 0.51 (to the nearest hundredth)

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If we square both sides of the equation we got:

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Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

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