Question

In the figure, $ABCD$ is a parallogram. The diagonals $AC$ and $BD$ intersect at $O$; and $\angle DAC={40}^{\circ },\angle CAB={35}^{\circ }$; and $\angle DOC={110}^{\circ }$. Calculate the $\angle ABO,\angle ADC,\angle ACB$, and $\angle CBD$

Answer 1

Data: ABCD is a parallelogram AD $\parallel$ BC and DC $\parallel$ AB.
The diagonals AC and BD intersect at O
$\angle DAC$ = ${40}^{\circ }$ $\angle CAB$ = ${35}^{\circ }$and $\angle DOC$ = ${110}^{\circ }$
To find : (1)$\angle ABO$ (2) $\angle ADC$ (3) $\angle ACB$ (4)$\angle CBD$
Proof:$\angle DAC$ + $\angle CAB$ = $\angle A$
${40}^{\circ }$ + ${35}^{\circ }$ = $\angle A$
$\angle A$ = ${75}^{\circ }$
$\angle C$ = $\angle A={75}^{\circ }$(Opposite angles of parallelogram are equal)
$\angle D$ = $\angle A$ = ${180}^{\circ }$ (Supplementary angles)
$\angle D$ = ${180}^{\circ }$ -${75}^{\circ }$= ${105}^{\circ }$
$\angle B$ = $\angle D$ = ${105}^{\circ }$ (Opposite angles of parallelogram are equal)
$\angle DOC$ = $\angle AOB$ = ${110}^{\circ }$
Vertically opposite angles)
In AOB, $\angle A$ + $\angle O$ + $\angle B$ = ${180}^{\circ }$
(Sum of all angles of a triangle is ${180}^{\circ }$ )
${35}^{\circ }$ + ${110}^{\circ }$ + $\angle B$ = ${180}^{\circ }$
$\angle B$ = ${180}^{\circ }$ -${145}^{\circ }$
(1) $\angle ABO$ = ${35}^{\circ }$
(2) $\angle ADC$ = ${105}^{\circ }$(Proved)
(3) $\angle ACB$ = $\angle CBD$ = ${40}^{\circ }$
(Alternate angles, AD || BC)
$\angle CBD$ = ${105}^{\circ }$ -${35}^{\circ }$
(4) $\angle CBD$ = ${70}^{\circ }$
(1) $\angle ABO$ = ${35}^{\circ }$
(2) $\angle ADC$ = ${105}^{\circ }$
(3) $\angle ACB$ = ${40}^{\circ }$
(4)$\angle CBD$ = ${70}^{\circ }$