Posted on - 11-02-2017

Bank PO

In the figure given below, PQ is parallel to RS. ∠ABQ = 60°, ∠BCR = ? .

∠PBA + ∠ABQ = 180°

(Angle on straight line)

∠PBA + 60 = 180°

\ ∠PBA = 120°

∠BCR = ∠PBA = 120°

(Corresponding angles). .

In the given figure, AX || CY. What is the measurement of ∠ABP? .

By construction, draw line BZ, parallel to AX or CY. .

a + 120° = 180° and b + 130° = 180° (Interior angles on the same side of the transversal are supplementary)

\ a = 60° and b = 50°

a + b = 110°

110° + ∠ABP = 180°

&⇒ ∠ABP = 70°

In the following figure, in triangle ABC

∠A : ∠B : ∠ACB = 2 : 3 : 4. What is ∠ACD? .

∠ACB = 4/9 x 180° = 80°

∠ACB + ∠ACD = 180°

(Angle on a straight line)

∠ACD = 180° - 80°

\ ∠ACD = 100°

I is the incentre of DABC such that, ∠AIC = 120° and ∠BIC = 140°. Then ∠ACB = .

Given that I is the incentre of -ABC, and ∠AIC = 120° ∠BIC = 140°

&⇒ ∠AIB = 360° — (∠AIC + ∠BIC)

&⇒ ∠AIB = 360° — (120° + 140°)

&⇒ ∠AIB = 100°

In -AIB

∠IAB + ∠AIB + ∠ABI = 180°

&⇒ x + 100° + z = 180°

&⇒ x + z = 180° - 100° = 80° (1)

In -ABC

∠BAC + ∠ABC + ∠BCA = 180°

&⇒ 2x + 2z + 2y = 180°

&⇒ (x + z) + y = 90°

&⇒ from (1)

80° + y = 90°

&⇒ y = 10°

\ ∠ACB = 2(10°) = 20°

The external bisectors of ∠A and ∠C of a DABC intersect at 0. If ∠AOC = 80° then ∠ABC is .

Given that AO and CO are the external bisectors of ∠A and ∠C

Let ∠XAC = x and ∠YCA = y

&⇒ ∠OAC = x/2, ∠OCA = y/2

In -AOC, ∠OAC + ∠OCA + ∠AOC = 180°

&⇒ = 180° -
80° = 100°

&⇒ x + y = 200° _______(1)

\ ∠CAB = 180° - x,

∠ACB = 180° - y

In -ABC

∠ABC + ∠CAB + ∠ACB = 180°

&⇒ ∠ABC + (180 - x) + (180 - y) = 180°

&⇒ ∠ABC = 180° + (x + y) - 360°

&⇒ ∠ABC = 180° + 200° - 360

&⇒ ∠ABC = 20°

In the above parallelogram ABCD, ∠A = x + 30° and ∠D = x - 40°, what is the measure of ∠DCB?

In a parallelogram, sum of adjacent angles is equal to 180°

**\**** **x + 30 + x – 40 = 180

&⇒ x = 95°

∠DAB = x + 30 = 95 + 30 = 125°

\ ∠DCB = ∠DAB = 125°

(opposite angles of a parallelogram are equal)

In the above diagram ‘O’ is the center of circle and ∠ACB = 30°. What is angle ∠AOB? .

Angle made by an are at the center of the circle, is twice the angle made by it in the remaining part of the circle. .

\∠AOB = 2(30°) = 60°

In the above figure, AB and AD are tangents to the circle with centre O. C is a point on the circle. If ∠BAD = 40°, the measure of ∠BCD is.

Given,

AB and AD are tangents drawn form A to the circle with centre O. .

We now join OB and OD, which are the radii of the circle. .

In quadrilateral ABOD,

∠ABO = ∠ADO = 90° (Radius makes an angle of 90° with the tangent at the point of tangency)

\ ∠BOD = 180°- ∠BAD

= 180° - 40° = 140°

As ∠BCD = 1/2(∠BOD)°

= 1/2(140)° = 70°

\ ∠BCD = 70°

AB and BC are the two chords of a circle having equal lengths. ∠ACB = ∠OBC where O is centre of the circle. Find the radius of the circle, if AB = 2 cm. .

Given AB = BC \ ∠ACB = ∠BAC (In a circle, angles made by equal chords are equal)

Let ∠ACB = q

\ ∠ACB = ∠BAC = ∠OBC = q

\ ∠AOB = 2q, ∠ABC = 180 - 2q

\ ∠ABO = 180 - 3q

∠AOB + ∠ABO + ∠OAB = 180°

\ 2q + 180° - 3q + ∠OAB = 180°

\ ∠OAB = q. .

\ ‘O’ is on AC, i.e., AC is diameter. .

\ AC = 2r = =

\ r = cm.

In the figure given below ABC and PBC are two triangles which are inscribed in a circle. If ∠A = 60° then find ∠P. .

Given that -ABC and -PBC are two triangles inscribed in a circle. .

The given triangles have same base. .

\ ∠P and ∠A are the angles in the same segment. .

Hence ∠A = ∠P = 60°