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
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.
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)
It is 7.5 cm.
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
A one-percent sector subtends an angle of 3.6 degrees at the centre.
The answer will depend on the location of the points B, P and C.
what is the measurement in degrees of angle 7
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
For a circle: Arc Length= R*((2*P*A)/(360)) R being radius, P being pi (3.14159), and A being the measure of the central angle.
If you mean a circle inscribed in the square: C = circumference, π = pi, r = radius, s = side, P = perimeter C = 2πr r = s/2 C = πs s = P/4 C = πP/4 So for this problem, the circumference is 9π, or about 28.3 If you mean a square inscribed in the circle, the computations are practically the same, except: r = sqrt(2)s/2 C = sqrt(2)πs C = sqrt(2)πP/4 So for this problem, the circumference is sqrt(2)9π, or about 40.0
The P-P-P bond angle in white phosphorus is approximately 60 degrees.
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.
I presume you mean the circumference of a circle. If P and Q are two points on the circumference of a circle with center O, the number of degrees in the arc PQ is defined as the number of degrees in the angle POQ.
the f-p-f bond angle is 120the cl -p-cl bond angle is 180and the f - p - cl bond angle is 90
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)
Electrical angle is half of mechanical angle in unipolar electrical machines. In multipolar electrical machines, the relationship between the mechanical angle and electrical angle is Electrical angle = (P/2) x Mechanical angle where: P = Number of poles.