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malfutka [58]
3 years ago
15

Laws of Sines with find the angle.

Mathematics
2 answers:
Fiesta28 [93]3 years ago
5 0

Answer:

1. H ≈ 49.1°

2. F ≈ 38.0°

3. R ≈ 17.0°

Step-by-step explanation:

1.

We need to find angle H, which is the third angle. We are given two sides and an angle, so we can use the Law of Sines to solve.

The Law of Sines states that for a triangle with side lengths a, b, and c and angles A, B, and C:

\frac{a}{sinA} =\frac{b}{sinB}  =\frac{c}{sinC}

Here, we can say that a = 22, b = 27, and B = 68. We want to find "A", which is the same angle as H:

\frac{a}{sinA} =\frac{b}{sinB}

\frac{22}{sinH} =\frac{27}{sin(68)}

Solve for H:

H ≈ 49.1°

2.

Use the Law of Sines again (use the same method as above):

\frac{17}{sin(108)} =\frac{11}{sin(F)}

Solve for F:

F ≈ 38.0°

3.

Use the Law of Sines again (use the same method as before):

\frac{9}{sin(26)} =\frac{6}{sin(R)}

Solve for R:

R ≈ 17.0°

Jobisdone [24]3 years ago
4 0
A divide by sin A = b divide by sin B
1. 27 divide by sin 68=22 divide by sin x
Sin x=sin 68 times 22 divide by 27
Sinx=0.65696
X=71.7 degrees
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In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track
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Approximately 0.31\; \rm m, assuming that g = 9.81\; \rm N \cdot kg^{-1}.

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned} .

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\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}.

Normal force between the sled and the slope:

\begin{aligned}F_{\rm N} &= W\cdot  \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}.

Calculate the kinetic friction between the sled and the slope:

\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}.

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\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}.

Distance travelled along the slope:

\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}.

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}.

In other words, the sled and its passenger would have lost (approximately) ((137 + 68.1)\, h)\; {\rm J} of energy when it is at a height of h\; {\rm m}.

The initial amount of energy that the sled and its passenger possessed was \text{KE} = 63\; {\rm J}. All that much energy would have been converted when the sled is at its maximum height. Therefore, when h\; {\rm m} is the maximum height of the sled, the following equation would hold.

((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}.

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