In a parallelogram $ \mathrm{ABCD}, \angle \mathrm{A}: \angle \mathrm{B}=2: 3, $ then find angle $ \mathrm{D} $.
Given:
In a parallelogram \( \mathrm{ABCD}, \angle \mathrm{A}: \angle \mathrm{B}=2: 3 \).
To do:
We have to find angle \( \mathrm{D} \).
Solution:
We know that,
Sum of the angles in a parallelogram is $360^o$ and opposite angles are equal.
Let $\angle A=2x$ and $\angle B=3x$.
This implies,
$\angle C=\angle A=2x$ and $\angle D=\angle B=3x$.
Therefore,
$2x+3x+2x+3x=360^o$
$10x=360^o$
$x=\frac{360^o}{10}$
$x=36^o$
$\Rightarrow 3x=3(36^o)=108^o$
Therefore, the measure of $\angle D$ is $108^o$.
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