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Reika [66]
2 years ago
10

I need help quickly, I'm preparing for a test! Help is appreciated!

Mathematics
1 answer:
wel2 years ago
6 0

The measure of angle ∠A and angle ∠C will be 65.37° and 17.63°. And the measure of the length of CA will be 29.48 inches.

<h3>What is trigonometry?</h3>

The connection between the lengths and angles of a triangular shape is the subject of trigonometry.

In triangle ABC, the side AB is 9 in, the side BC is 27 in, and the angle B is 97 degrees.

Then the other two angles and the side CA will be

The side CA is given by the cosine rule. Then we have

\rm CA^2 = AB ^2+ BC^2 - 2 \times AB \times BC \cos B\\\\\\CA^2 = 9^2 + 27^2 - 2 \times 9 \times 27 \cos 97^o\\\\\\CA^2 = 810 + 59.23\\\\\\CA ^2 = 869.23\\\\\\CA \ = 29.48 \  in

Then angle A will be given by the sine rule. Then we have

\dfrac{27}{\sin A} = \dfrac{29.48}{\sin 97^0}\\\\\\\sin A = \dfrac{27 \sin 97^0}{29.48}\\\\\\\sin A = 0.9090\\\\\\A  \ \  \ \ = 65.37^0

Then the angle C will be

     ∠A + ∠B + ∠C = 180°

65.37° + 97° + ∠C = 180°

                       ∠C = 17.63°

More about the trigonometry link is given below.

brainly.com/question/22698523

#SPJ1

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The wind has a speed of 35 mi/h and a direction S-10°-E.

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\begin{gathered} x=560\cdot\sin (20\degree)+35\cdot\sin (10\degree) \\ x\approx560\cdot0.342+35\cdot0.174 \\ x\approx191.53+6.08 \\ x\approx197.61 \end{gathered}

and the y-coordinate (S-N axis) is:

\begin{gathered} y=560\cdot\cos (20\degree)-35\cdot\cos (10\degree) \\ y\approx560\cdot0.940-35\cdot0.985 \\ y\approx526.23-34.47 \\ y\approx491.76 \end{gathered}

Then, the actual speed vector is v3=(197.61, 491.76).

The starting location for the third leg is R2=(216.66, 167.67) [taken from the previous answer].

Then, we have to calculate the displacement in 20 minutes using the actual speed vector.

We can calculate the movement in each of the axis. For the x-axis:

\begin{gathered} R_{3x}=R_{2x}+v_{3x}\cdot t \\ R_{3x}=216.66+197.61\cdot\frac{1}{3} \\ R_{3x}=216.66+65.87 \\ R_{3x}=282.53 \end{gathered}

NOTE: 20 minutes represents 1/3 of an hour.

We can do the same with the y-coordinate:

\begin{gathered} R_{3y}=R_{2y}+v_{3y}\cdot t \\ R_{3y}=167.67+491.76\cdot\frac{1}{3} \\ R_{3y}=167.67+163.92 \\ R_{3y}=331.59 \end{gathered}

The final position is R3 = (282.53, 331.59).

To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:

The distance can be calculated as if it was a right triangle:

\begin{gathered} d^2=x^2+y^2_{} \\ d^2=282.53^2+331.59^2 \\ d^2=79823.20+109951.93 \\ d^2=189775.13 \\ d=\sqrt[]{189775.13} \\ d\approx435.63 \end{gathered}

The angle, from E to N, can be calculated as:

\begin{gathered} \tan (\alpha)=\frac{y}{x} \\ \tan (\alpha)=\frac{331.59}{282.53} \\ \tan (\alpha)\approx1.1736 \\ \alpha=\arctan (1.1736) \\ \alpha=49.56\degree \end{gathered}

If we want to express it from N to E, we substract the angle from 90°:

\beta=90\degree-\alpha=90-49.56=40.44\degree

Answer: the final location can be represented with the vector (282.53, 331.59).

1) The distance from the origin is 435.63 miles and

2) the direction is N-40°-E.

7 0
1 year ago
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