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Lines and Angles

In the adjoin...

Question

In the adjoining figure AB || CD and PQ, QR intersects AB ad CD both at E, F and G, H respectively. Given that $m ∠ P E B = 80^{\circ}$ , $m ∠ Q H D = 120^{\circ}$ and $m ∠ P Q R = x^{\circ}$ , find the value of x:

40^{\circ}

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20^{\circ}

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100^{\circ}

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30^{\circ}

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Solution

The correct option is B $20^{\circ}$
$∠ P G H = 80^{\circ} \Rightarrow ∠ Q G H = 100^{\circ}$
$∠ Q H D = 120^{\circ} \Rightarrow ∠ C H Q = 60^{\circ}$
$∴ ∠ x + 100 + 60 = 80^{\circ}$
$\Rightarrow x = 20^{\circ}$

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In the adjoining figure AB || CD and PQ, QR intersects AB ad CD both at E, F and G, H respectively. Given that m∠PEB=80∘, m∠QHD=120∘ and m∠PQR=x∘, find the value of x:

The correct option is B 20∘∠PGH=80∘⇒∠QGH=100∘ ∠QHD=120∘⇒∠CHQ=60∘ ∴∠x+100+60=80∘ ⇒x=20∘

In the adjoining figure AB || CD and PQ, QR intersects AB ad CD both at E, F and G, H respectively. Given that $m ∠ P E B = 80^{\circ}$ , $m ∠ Q H D = 120^{\circ}$ and $m ∠ P Q R = x^{\circ}$ , find the value of x:

The correct option is B $20^{\circ}$
$∠ P G H = 80^{\circ} \Rightarrow ∠ Q G H = 100^{\circ}$
$∠ Q H D = 120^{\circ} \Rightarrow ∠ C H Q = 60^{\circ}$
$∴ ∠ x + 100 + 60 = 80^{\circ}$
$\Rightarrow x = 20^{\circ}$