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 parallelogramConsider 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 $$\angle B=\angle D={91}^{o}$$MathematicsRS AgarwalStandard IX

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