Q26 of 77 Page 16

In a rhombus ABCD, if ∠ACB =40°, then ∠ADB =

The diagonals in a rhombus are perpendicular,

So,

∠BPC = 90^o

From triangle BPC,

The sum of angles is 180°

So,

∠CBP = 180^o – 40^o – 90^o

= 50°

Since, triangle ABC is isosceles

We have,

AB = BC

So,

∠ACB = ∠CAB = 40^o

Again from triangle APB,

∠PBA = 180^o – 40^o – 90^o

= 50^o

Again, triangle ADB is isosceles,

So,

∠ADB = ∠DBA = 50^o

∠ADB = 50^o

The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD =50°, then ∠DPC =

ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC =35°, then ∠ABC =

In Δ ABC, ∠A =30°, ∠B=40° and ∠C=110°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are

The diagonals of a parallelogram ABCD intersect ay O. If ∠BOC =90° and ∠BDC=50°, then ∠OAB =