In a parallelogram ABCD, $∠ A = x^{o}, ∠ B = 3 x + 20^{o}$ , find $x$ .

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Solution

$A n y t w o a d j a c e n t a n g l e s o f a p a r a l l e l o g r a m a r e s u p p l e m e n t a r y . S o ∠ A + ∠ B = 180^{o} \Rightarrow x + 3 x + 20 = 180 \Rightarrow 4 x = 160 \Rightarrow x = 40 S o ∠ A = 40^{o}, ∠ B = 3 \times 40 + 20 = 140^{o} O p p o s i t e a n g l e s o f a p a r a l l e l o g r a m a r e e q u a l . S o ∠ C = ∠ A = 40^{o}, ∠ D = ∠ B = 140^{o}$

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