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 Fig., below, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove that AD = DP, PC = BC and DC = 2AD.
In a parallelogram ABCD, determine the sum of angles ∠C and ∠D .
In a parallelogram ABCD, if `∠`B = 135°, determine the measures of its other angles .
ABCD is a square. AC and BD intersect at O. State the measure of ∠AOB.
P and Q are the points of trisection of the diagonal BD of a parallelogram AB Prove that CQ is parallel to AP. Prove also that AC bisects PQ.
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following case, PQRS is a parallelogram?
∠P = 100°, ∠Q = 80°, ∠R = 95°
P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. If AD = 10 cm, then CD =
In a quadrilateral ABCD, ∠A + ∠C is 2 times ∠B + ∠D. If ∠A = 140° and ∠D = 60°, then ∠B=
In the given Figure, if AB = 2, BC = 6, AE = 6, BF = 8, CE = 7, and CF = 7, compute the ratio of the area of quadrilateral ABDE to the area of ΔCDF. (Use congruent property of triangles)
Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.
