Question

P,Q,R and S are points on the circumference of a circle, centre, O. TU is a tangent to the circle at the point S. Angle ROS =64\deg and angle QSU =58\deg . (i) Find the size of angle: (a) OSQ (b) SQR (c) QPS (d) QRS (b) QSR

Answer 1

(a)
`\( \angle OSQ = 122^\circ \)`

(b)
`\( \angle SQR = 122^\circ \)`

(c)
`\( \angle QPS = 64^\circ \)`

(d)
`\( \angle QRS = 58^\circ \)`

(e)
`\( \angle QSR = 64^\circ \)`

Step-by-step explanation:

In a circle, the angle subtended by an arc at the center is twice the angle subtended by it at any point on the remaining part of the circle. Using this property, we can find the angles:

(a)
`\( \angle OSQ = 2 * \angle ORS = 2 * 64^\circ = 128^\circ \):`:

The angle subtended by the arc OS at the center O is twice the angle
`\angle ROS` at any point on the remaining part of the circle.

(b)
`\( \angle SQR = \angle OSQ = 128^\circ \)`:

This is because
`\angle SQR` and
`\( \angle OSQ \)` are vertically opposite angles, and in a circle, vertically opposite angles are equal.

(c)
`\( \angle QPS = (1)/(2) * \angle QOS = (1)/(2) * 64^\circ = 32^\circ \):`

The angle
`\( \angle QPS \)` is half of the angle subtended by the arc QOS at the center O.

(d)
`\( \angle QRS = \angle QSU = 58^\circ \):`

This is because
`\( \angle QRS \)` and
`\( \angle QSU \)` are alternate angles formed by the tangent TU and the chord RS.

(e)
`\( \angle QSR = \angle QOS - \angle ROS = 64^\circ - 58^\circ = 6^\circ \):`

This is because
`\( \angle QSR \)` is the difference between the angle subtended by the arc QOS and the angle
`\( \angle ROS \)` at point S.

Answer 2

Using properties of circles and tangents, we found the sizes of various angles, which involve concepts such as the angle at the center being twice the angle at the circumference, alternate segment theorem, the sum of opposite angles in a cyclic quadrilateral, and the sum of angles in a triangle.

Step-by-step explanation:

The question involves finding the sizes of various angles in a circle with a tangent drawn from a point on its circumference. Given that angle ROS is 64 degrees and angle QSU is 58 degrees, we can use the properties of circles to determine the other angles.

(a) To find the angle OSQ, we use the fact that the angle at the center is twice the angle at the circumference. Therefore, if the angle ROS is 64 degrees, the angle at the circumference, OSQ, is 32 degrees.

(b) Angle SQR is related to angle QSU by the alternate segment theorem, which states that the angle between the tangent and the chord is equal to the angle in the alternate segment. Therefore, the angle SQR is 58 degrees.

(c) Angle QPS is the sum of angles OSQ and SQR since they form a straight line from point S. Therefore, QPS is 32 degrees + 58 degrees = 90 degrees.

(d) To find angle QRS, we need to recognize that it is a cyclic quadrilateral and the sum of opposite angles in a cyclic quadrilateral is 180 degrees. Since angle QPS is 90 degrees, angle QRS is 180 degrees - 90 degrees = 90 degrees as well.

(e) Angle QSR is the remaining angle in triangle QRS, where the sum of angles in a triangle is 180 degrees. Given that angle QPS is 90 degrees and angle SQR is 58 degrees, QSR is 180 degrees - 90 degrees - 58 degrees = 32 degrees.