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Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center?
Draw a circle with centre O. draw a tangent PR touching circle at P. Draw QP perpendicular to RP at point P, Qp lies in the circle. Now, angle OPR = 90 degree (radius perpendicular to tangent) also angle QPR = 90 degree (given) Therefore angle OPR = angle QPR. This is possible only when O lies on QP. Hence, it is prooved that the perpendicular at the point of contact to the tangent to a circle passes through the centre. Answer By- Rajendra Meena, Jaipur, India. email: rajendra.meena21@gmail.com
How sine is the y coordinate of a unit circle?
Sine is NOT the y coordinate: it is the sine of the angle made by the x-axis and the radius from a point on the circle. It is the cosine of the angle made with the y-axis.Consider any point, P, on the unit circle with coordinates (x, y). And let Q be the foot of the perpendicular from P to the x-axis. Then y = PQ.Now, in the right angled triangle OPQ, if OP makes an angle theta with the x axis, then sin(theta) = PQ/OP = y/OP and since OP is the radius of a unit circle, OP = 1 so that sin(theta) = y.
An angle of 50 degrees in standard position has its terminal side in the first quadrant. What are the coordinates of point P on the unit circle?
Draw a picture of this you will see that you can form a triangle with the radius of the circle being 1. This is the hypotenus of the triangle Using trig, x = cos 50 y = sin 50 P is (x,y) = P is (cos 50,sin 50) = (0.643, 0.766)
In circle P XY 2.4 cm. What is the circumference of the circle?
It is 7.5 cm.
If angle P is three less than twice the measure angle Q angle P and angle Q are supplementary angles what are their measurements?
Assuming the angles are expressed in degrees: P = 2Q -3° (because "angle P is three less than twice the measure angle Q") P + Q = 180° (because they are supplementary angles) P+Q = 2Q - 3° + Q = 3Q -3° = 180° 3Q = 183° Q = 61° P = 2∙61° -3° = 122° - 3° = 119° If the angles are expressed in radians, the math is similar except you start with P = 2Q - 3 and P+Q = π yielding P = 2π/3 -1 and Q = π/3 +1
