trigonometric functions/ radians
the angle Ə is in the first quadrant, and cos Ə = 2/√5
a) draw a diagram to show the angle in standard position and a point P on its terminal arm
b) determine possible coordinates for P
We can't draw diagrams here, but theta is the angle whose tangent is 1/2, which is 26.56 dgrees, and the coordinates of P are (2,1). The other end of the hypotenuse is at (0,0)
To draw a diagram showing the angle in standard position and a point P on its terminal arm, follow these steps:
a)
1. Start by drawing the x-axis (horizontal) and y-axis (vertical) intersecting at the origin point (0, 0).
2. Locate the first quadrant, which lies in the top right area of the coordinate plane.
3. Draw a ray originating from the origin towards the first quadrant. This ray represents the initial side of the angle.
4. Mark a point, which we'll call P, somewhere on this ray to represent the terminal arm of the angle.
b)
To determine the possible coordinates for point P, we need to use the given information that cos Ə = 2/√5. Since cosine is the ratio of the adjacent side to the hypotenuse in a right triangle, we can set up a right triangle with the adjacent side being 2 and the hypotenuse being √5.
1. Let point A represent the intersection of the ray and the unit circle.
2. Draw a horizontal line from point A to the y-axis to form a right triangle.
3. Label the right angle of the triangle as angle Ə.
4. Mark the adjacent side of the triangle with a length of 2 and the hypotenuse with a length of √5.
5. Now, we can find the length of the opposite side using the Pythagorean theorem:
a² + b² = c²
(opposite side)² + 2² = (√5)²
(opposite side)² + 4 = 5
(opposite side)² = 1
opposite side = 1 or -1 (since we are in the first quadrant and opposite side is positive)
From this, we have two sets of possible coordinates for point P:
1. P(2, 1): When the opposite side is 1, the coordinates of point P would be (2, 1) in the first quadrant.
2. P(2, -1): When the opposite side is -1, the coordinates of point P would be (2, -1). However, since the angle is in the first quadrant, we can disregard this solution.
Therefore, the possible coordinates for point P are (2, 1) in the first quadrant.