In the adjoining figure, ABCD is a parallelogram in which =35^ , =40^ and =150^ . Calculate i. , ii. , iii. , iv. .

ABCD is a parallelogram i. = =105^ [Vertical opposite angels] Now in , we have + + =180^ 35^ +105^ + =180^ 140^ + =180^ =180^ -140^ =40^ . ii. Since AB | DC and BD is a transversal So, = [alternate angles] = = = =40^ =40^ iii. As AB | CD and AC is a transversal So, = =40^ [alternate opposite angles] iv. = - But, + + + =360^ [ ABCD is a parrellogram] 2 +2 =360^ 2 (40^ +35^ )+2 =360^ 150^ +2 =360^ 2 =360^ -150^ =210^ = 210^ 2 =105^ and = - =105^ -40^ =65^ =65^

In the adjoining figure, ABCD is a parallelogram in which =35^ , …

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Question

In the adjoining figure, $\text{ABCD}$ is a parallelogram in which $\angle\text{BAO}=35^{\circ},\angle\text{DAO}=40^{\circ}$ and $\angle\text{COD}=150^{\circ}.$ Calculate
$i. \angle\text{ABO},$
$ii. \angle\text{ODC},$
$iii. \angle\text{ACB},$
$iv. \angle\text{CBD}.$

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Answer

$\text{ABCD}$ is a parallelogram

$i. \angle\text{AOB}=\angle\text{COB}=105^{\circ} [$Vertical opposite angels$]$
Now in $\triangle\text{AOB},$ we have
$\angle\text{OAB}+\angle\text{AOB}+\angle\text{ABO}=180^{\circ}$
$\Rightarrow35^{\circ}+105^{\circ}+\angle\text{ABO}=180^{\circ}$
$\Rightarrow140^{\circ}+\angle\text{ABO}=180^{\circ}$
$\Rightarrow\angle\text{ABO}=180^{\circ}-140^{\circ}=40^{\circ}.$
$ii.$ Since $AB\ \|\ DC$ and $BD$ is a transversal
So, $\angle\text{ABD}=\angle\text{CDB} [$alternate angles$]$
$\Rightarrow\angle\text{CDO}=\angle\text{CDB}=\angle\text{ABD}=\angle\text{ABO}=40^{\circ}$
$\therefore\angle\text{ODC}=40^{\circ}$
$iii.$ As $AB\ \|\ CD$ and $AC$ is a transversal
So, $\angle\text{ACB}=\angle\text{DAC}=40^{\circ} [$alternate opposite angles$]$
$iv. \angle\text{CBD}=\angle\text{B}-\angle\text{ABO}$
But, $\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^{\circ}[\because
\text{ABCD}$ is a parrellogram$]$
$\Rightarrow2\angle\text{A}+2\angle\text{B}=360^{\circ}$
$\Rightarrow2\times(40^{\circ}+35^{\circ})+2\angle\text{B}=360^{\circ}$
$\Rightarrow150^{\circ}+2\angle\text{B}=360^{\circ}$
$\Rightarrow2\angle\text{B}=360^{\circ}-150^{\circ}=210^{\circ}$
$\Rightarrow\angle\text{B}=\frac{210^{\circ}}{2}=105^{\circ}$
and $\angle\text{CBD}=\angle\text{B}-\angle\text{ABO}$
$=105^{\circ}-40^{\circ}=65^{\circ}$
$\angle\text{CBD}=65^{\circ}$