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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Updated on: 10-Oct-2022

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