Advertisements
Advertisements
Question
In the given figure, ∠A = 64°, ∠ABC = 58°. If BO and CO are the bisectors of ∠ABC and ∠ACB respectively of ΔABC, find x° and y°
Advertisements
Solution
In the given ΔABC
∠A = 64° and ∠B = 58°
∠C = 180° − (64° + 58°)
= 180° – 122°
= 58°
Since OC is the bisector of ∠C
y = `(58^circ)/2`
= 29°
Given ΔOBC
∠OCB = `(58^circ)/2` = 29°
∠OCB = 29°
∴ ∠BOC = 180° − (29° + 29°)
x = 180° – 58°
x = 122°
∠x = 122° and ∠y = 29°.
APPEARS IN
RELATED QUESTIONS
In a parallelogram ABCD, determine the sum of angles ∠C and ∠D .
The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.
In Fig. below, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ΔABC.
Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram

In a parallelogram ABCD, if ∠A = (3x − 20)°, ∠B = (y + 15)°, ∠C = (x + 40)°, then find the values of xand y.
In a parallelogram ABCD, the bisector of ∠A also bisects BC at X. Find AB : AD.
We get a rhombus by joining the mid-points of the sides of a
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
ABCD is a square, diagonals AC and BD meet at O. The number of pairs of congruent triangles with vertex O are
Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.
