- ACS -110- then - BAC -A- 40-B- 70-C- 110-D- 30-

The correct option is A 40° ∠ PAC =∠ ACS = 110^∘ ∠ BAC =∠ PAC ∠ PAB = 110^∘ 70^∘ = 40^∘ ...

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Byju's Answer

Standard IX

Mathematics

Lines Parallel to the Same Line

∠ ACS =110∘, ...

Question

In the given figure PQ || RS $∠ P A B = 70^{\circ}$ , $∠ A C S = 110^{\circ}$ , then $∠ B A C =$

40°

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70°

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110°

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30°

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Solution

The correct option is A
40°

$∠ P A C = ∠ A C S = 110^{\circ}$

$∠ B A C = ∠ P A C - ∠ P A B = 110^{\circ} - 70^{\circ} = 40^{\circ}$

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In the given figure PQ || RS ∠PAB=70∘ , ∠ACS=110∘ , then ∠BAC=

The correct option is A 40° ∠PAC=∠ACS=110∘ ∠BAC=∠PAC−∠PAB=110∘−70∘=40∘

In the given figure PQ || RS $∠ P A B = 70^{\circ}$ , $∠ A C S = 110^{\circ}$ , then $∠ B A C =$

The correct option is A
40°

$∠ P A C = ∠ A C S = 110^{\circ}$

$∠ B A C = ∠ P A C - ∠ P A B = 110^{\circ} - 70^{\circ} = 40^{\circ}$