example.1
In the
figure given below, PQ is parallel to RS. ∠ABQ = 60°, ∠BCR = ? .
Solution
∠PBA + ∠ABQ = 180°
(Angle on straight line)
∠PBA + 60 = 180°
\ ∠PBA = 120°
∠BCR = ∠PBA = 120°
(Corresponding angles). .
example.2
In the given figure, AX || CY. What is the
measurement of ∠ABP? .
Solution
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°
example.3
In the
following figure, in triangle ABC
∠A : ∠B : ∠ACB = 2 : 3 : 4.
What is ∠ACD? .
Solution
∠ACB = 4/9 x 180° = 80°
∠ACB + ∠ACD = 180°
(Angle on a straight line)
∠ACD = 180° - 80°
\ ∠ACD = 100°
example.4
I is the incentre of DABC such that, ∠AIC = 120° and ∠BIC = 140°. Then
∠ACB = .
Solution
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°
example.5
The external bisectors of ∠A and ∠C of a DABC
intersect at 0. If ∠AOC = 80° then ∠ABC is .
Solution
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°
example.6
In the above parallelogram ABCD, ∠A = x + 30° and ∠D = x - 40°,
what is the measure of ∠DCB?
Solution
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)
example.7
In the above diagram ‘O’ is the center of
circle and ∠ACB = 30°. What is angle ∠AOB? .
Solution
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°
example.8
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.
Solution
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°
example.9
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. .
Solution
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.
example.10
In the figure given below ABC and PBC are two
triangles which are inscribed in a circle. If ∠A = 60° then find ∠P. .
Solution
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°