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Question

In a parallelogram $$ABCD$$, if $$\angle A={(2x+5)}^{o}$$ and $$\angle B={(3x-5)}^{o}$$, find the value of $$x$$ and the measure of each angle of the parallelogram.


Solution

We know that the opposite angles are equal in a parallelogram

Consider parallelogram $$ABCD$$

So we get

$$\angle A=\angle C={ \left( 2x-25 \right)  }^{ o }$$

$$\angle B=\angle D={ \left( 3x-5 \right)  }^{ o }$$

We know that the sum of all the angles of a parallelogram is $${360}^{o}$$

so it can be written as

$$\angle A+\angle B+\angle C+\angle D={360}^{o}$$

By subtituting the values in the above equation

$$(2x+25)+(3x-5)+(2x+25)+(3x-5)={360}^{o}$$

by addition we get

$$10x+{40}^{o}={360}^{o}$$

$$10x={320}^{o}$$

$$x={32}^{o}$$

now substituting the value $$x$$

$$\angle A=\angle C={(2x+25)}^{o}={(2(32)+25)}^{o}$$

$$\angle A=\angle C={(54+25)}^{o}$$

by addition

$$\angle A=\angle C={89}^{o}$$

$$\angle B=\angle D={(3x-5)}^{o}={(3(32)-5)}^{o}$$

$$\angle B=\angle D={(96-5)}^{o}$$

by subtraction

$$\angle B=\angle D={91}^{o}$$

therefore $$x={32}^{o}$$, $$\angle A=\angle C={89}^{o}$$ and $$\angle B=\angle D={91}^{o}$$

Mathematics
RS Agarwal
Standard IX

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