Question

The diagram shows a circle, centre 0. VT is a diameter and ATB is a tangent to the circle at T. U, V, W, X lie on the circle and angle VOU = 70 degrees. Calculate e, f, g, h.

Answer 1

Answer:

☛ e = 35°

☛ f = 55°

☛ g = 55°

☛ h = 125°

Explanation:

Given:

☛ VT is a diameter and ATB is a tangent at T.

☛ U, V, W, X lie on the circle and ∠VOU = 70°

To Find:

☛ E, F, G, H

Solution:

Here,

∠VOU + ∠TOU = 180° ( Linear Pair )

☛ ∠TOU = 180° - 70° { given }

☛ ∠TOU = 110° ...(1)

∠TOU = 2 ∠TWU

∵ Angle subtended by an arc at the center is double the angle subtended anywhere on the circle.

☛ 110° = 2 × g { from (1) }

☛ g = 55° ...(2)

Now,

In △TOU,

☛ ∠TWU + ∠TXU = 180°

∵ Sum of opposite angles of a cyclic quadrilateral ( Here, TXUW ) is 180°

☛ g + h = 180°

☛ h = 180° - 55° { from (2) }

☛ h = 125° ...(3)

Again,

∠TWV = 90°

∵ Angle subtended by the diameter on the circumference is 90°

☛ ∠TWV = ∠TWU + ∠VWU

☛ 90° = g + ∠VWU

☛ ∠VWU = 90° - 55° { from (2) }

☛ ∠VWU = 35° ...(4)

Now,

∠VWU = ∠VTU ...(5)

∵ Angles subtended by an arc on the circle are equal to each other.

Angles = ∠VWU , ∠VTU

Subtend by the minor arc VU

☛ e = 35° ...(6)

∵ ∠VWU = 35° { from (4) }

and ∠VWU = ∠VTU { from (5) }

So,

∠VTB = 90°

∵ VTB is tangent and OT as radius

We know, Radius is perpendicular to the tangent at the point of contact.

☛ ∠VTB = ∠VTU + ∠UTB

☛ e + f = 90°

☛ f = 90° - 35° { from (6) }

☛ f = 55° ...(7)

Hence, From (2), (3), (6), (7)

g = 55° , h = 125° , e = 35° , f = 55°