Math  /  Trigonometry

QuestionThe terminal side of angle AA in standard position goes through the point P(5,2)P(-5,-2). Draw the ref triangle in the cartesian plane that is made by the point, by first plotting the point and then clickin all of the vertices of the triangle. If a trigonomic value is undefined then answer INF.
Clear All Draw: Polygon
Find the following: (round to 3 decimal places) sinA=cosA=tanA=A=\begin{array}{l} \sin A=\square \\ \cos A=\square \\ \tan A=\square \\ A=\square \end{array}
The reference angle = \square

Studdy Solution

STEP 1

1. The point P(5,2) P(-5, -2) lies on the terminal side of angle A A in standard position.
2. The angle A A is measured from the positive x-axis to the line segment connecting the origin to the point P(5,2) P(-5, -2) .
3. We are to find the trigonometric values of angle A A and the reference angle.

STEP 2

1. Plot the point and draw the reference triangle.
2. Calculate the hypotenuse of the reference triangle.
3. Determine the trigonometric ratios.
4. Find the angle A A and the reference angle.

STEP 3

Plot the point P(5,2) P(-5, -2) on the Cartesian plane. Draw a line from the origin (0,0)(0,0) to P(5,2) P(-5, -2) . This line is the terminal side of angle A A .
Draw a vertical line from P(5,2) P(-5, -2) to the x-axis at the point (5,0)(-5, 0). This forms a right triangle with vertices at (0,0)(0,0), (5,0)(-5, 0), and (5,2)(-5, -2).

STEP 4

Calculate the hypotenuse r r of the reference triangle using the distance formula: r=(5)2+(2)2=25+4=29 r = \sqrt{(-5)^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29}

STEP 5

Determine the trigonometric ratios:
sinA=oppositehypotenuse=2290.371\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{-2}{\sqrt{29}} \approx -0.371
cosA=adjacenthypotenuse=5290.928\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{-5}{\sqrt{29}} \approx -0.928
tanA=oppositeadjacent=25=0.4\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{-2}{-5} = 0.4

STEP 6

Find the angle A A using the inverse tangent function: A=tan1(0.4)21.801 A = \tan^{-1}(0.4) \approx 21.801^\circ
Since the point P(5,2) P(-5, -2) is in the third quadrant, the actual angle A A is: A=180+21.801=201.801 A = 180^\circ + 21.801^\circ = 201.801^\circ
The reference angle is the acute angle formed with the x-axis, which is: Reference angle=180201.801=21.801 \text{Reference angle} = 180^\circ - 201.801^\circ = 21.801^\circ
The trigonometric values and angles are:
sinA=0.371cosA=0.928tanA=0.4A=201.801Reference angle=21.801\begin{array}{l} \sin A = -0.371 \\ \cos A = -0.928 \\ \tan A = 0.4 \\ A = 201.801^\circ \\ \text{Reference angle} = 21.801^\circ \end{array}

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