Question

If measures opposite angles of a parallelogram are (60-x)° and (3x-4)°. Then find the measure of angles of a parallelogram​

Answer 1

$\huge\bf\red{\underline{\underline{Given}}}\::$

• ${\sf{\gray{Measures \:of\:opposite \:angles \:=\: }}}{\sf{\pink{(60-x)^{\circ}\:\&\:(3x-4)^{\circ}}}}$

$\huge\bf\red{\underline{\underline{To\:Find}}}\::$

• ${\sf{\blue{Measure\: of\:angles\:of\:the\: parallelogram}}}$

$\huge\bf\red{\underline{\underline{Solution}}}\::$

$\star\:\sf\underline\pink{In\: parallelogram,\:The\: opposite \:angles\:are\:equal}$

• $\sf\orange{\angle B \:=\: \angle D}$
• $\sf\orange{\angle A \:=\: \angle B}$

$\to\:\:\sf\purple{(60\:-\:x)^{\circ}\:=\:(3x\:-\:4)^{\circ}}$

$\to\:\:\sf\green{60^{\circ}\:-\:x\:=\:3x\:-\:4^{\circ}}$

$\to\:\:\sf\purple{60^{\circ}\:+\:4x^{\circ}}$

$\to\:\:\sf\green{x\:=\:\dfrac{\cancel{64}}{\cancel{4}}}$

$\to\:\:\sf\purple{x\:=\:16^{\circ}}$

$\star\:\sf\underline\red{Finding\:\angle A}$

$\:\:\:\hookrightarrow\:\sf\orange{60\:-\:16}$

$\:\:\:\hookrightarrow\:\sf\blue{\angle A \:=\:44^{\circ}}$

$\:\:\:\hookrightarrow\:\sf\orange{So,\:\angle C \:=\:44^{\circ}}$

$\star\:\sf\underline\red{Now,\:\angle A\:+\:\angle B\:=\:180^{\circ}}$

$\:\:\:\mapsto\:\sf\purple{44^{\circ}\:+\:\angle B\:=\:180^{\circ}}$

$\:\:\:\mapsto\:\sf\pink{\angle B\:=\:180^{\circ}\:-\:44^{\circ}}$

$\:\:\:\mapsto\:\sf\purple{\angle B\:=\:136^{\circ}}$

$\:\:\:\mapsto\:\sf\pink{So,\:\angle D\:=\:136^{\circ}}$

$\:\:\star\:\bf\underline\gray{Hence}\:\: \begin{cases}{\sf{\red{\angle A \:=\:44^{\circ}}}}\\ {\sf{\orange{\angle C\:=\:44^{\circ}}}}\\{\sf{\blue{\angle B \:=\:136^{\circ}}}}\\{\sf{\purple{\angle D \:=\:136^{\circ}}}}\end{cases}$