Question

Diagonals AC and BD of a parallelogram intersect at O. Of angle boc=90' and angle bdc=50',then find angle oab ​

Answer 1

Answer:

/ OAB = 40 degree

Step-by-step explanation:

[NOTE : PLEASE DRAW THE FIGURE ACCORDING TO THE GIVEN QUESTION SO AS TO UNDERSTAND THE ANSWER]

GIVEN : In parallelogram ABCD

diagonal AC and diagonal BD intersect at point O

angleBOC=90degree

angleBDC=50degree

TO FIND OUT: angle OAB

SOLUTION:

In parallelogram ABCD

line AB is parallel to line DC and diagonal BD is rhe transversal

angle BDC=angleDBA ... alternate angles

angle DBA=50 degree

angle OBA=50 degree ... (D-O-B)...(1)

m/ AOB+m/ BOC =180degree .. (linear pair

angles are

supplementary )

m/ AOB +90=180

m/ AOB =180-90

m/ AOB =90 degree ...(2)

In /\ AOB

m/ AOB +m/ OBA +m/ OAB =180 degree

90+50+m/ OAB =180

140 + m/ OAB =180

m / OAB =180-140

m / OAB =40 degree

Answer= the measure of angle OAB is 40 degree.