In ABC, B = 35 , C = 65 and the bisector of BAC meets BC in D. T – McqOptions

MCQOPTIONS

1.	In ABC, B = 35 , C = 65 and the bisector of BAC meets BC in D. Then ADB is :
A.	40
B.	75
C.	90
D.	105
Answer» E.

Discussion

No Comment Found

Related MCQs

In the adjoining figure, AB || CD, t is the traversal, EG and FG are the bisectors of BEE and DFE respectively, then EGF is equal to :
In the given figure, DE || BC and DE : BC = 3 : 5 the ratio of the areas of ADE and the trapezium BCED.
If H is the orthocentre of ABC, then the orthocentre of HBC is (fig. given) :
In a triangle ABC, the length of the sides AB, AC and BC are 3, 5 and 6 cm respectively. If a point D on BC is drawn such that the line AD bisects the angle A internally, then what is the length of BD?
X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P; CX and BY are joined intersecting in Q. Then PXQY is a :
ABCD is a rhombus with ABC = 56 , then ACD is equal to :
In the given figure, In a ABC , B = C. If AM is the bisector of BAC and AN BC, then MAN is equal to :
ABCD is a square, F is mid point of AB and E is a point on BC such that BE is one-third of BC. If area of FBE = 108 m2, then the length of AC is:
The diagonals of a rectangle ABCD meet at 0. If BOC = 44 , then OAD is equal to :
ABCD is a parallelogram P, Q, R and S are points on sides AB, BC, CD and DA respectively such that AP = DR. If the area of the parallelogram ABCD is 16 cm2, then the area of the quadrilateral PQRS is: