In the given figure, l || m and p is the transversal. If $∠ 1$ and $∠ 2$ are in the ratio 1 : 2, find the measure of the angle $7$ in degrees

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Solution

L1 & L2 are in ratio 1:2
$l e t L 1 = k$
$L 2 = 2 k$
$L 1 + L 2 = 180^{\circ} (∵ s u m o f a n g l e s o n a s t r a i g h t l i n e = 180^{\circ})$
$⟹ k + 2 k = 180^{\circ}$
$⟹ 3 k = 180^{\circ}$
$⟹ k = 60^{\circ}$
$⟹ L 1 = 60^{\circ}, L 2 = 120^{\circ}$
$L 2 + L 3 = 180^{\circ} (s u m o f a n g l e s o n a s t r a i g h t l i n e = 180^{\circ})$
$⟹ L 3 + 120^{\circ} = 180^{\circ}$
$⟹ L = 60^{\circ}$
$L 3 = L 7 (∵ l & m a r e p a r a l l e l, ∴ c o r r e s p o n d i n g a n g l e s)$
$L 7 = 60^{\circ}$

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In the given figure, l || m and p is the transversal. If ∠1 and ∠2 are in the ratio 1 : 2, find the measure of the angle 7 in degrees

L1 & L2 are in ratio 1:2letL1=kL2=2kL1+L2=180∘(∵sumofanglesonastraightline=180∘)⟹k+2k=180∘⟹3k=180∘⟹k=60∘⟹L1=60∘,L2=120∘L2+L3=180∘(sumofanglesonastraightline=180∘)⟹L3+120∘=180∘⟹L=60∘L3=L7(∵l&mareparallel,∴correspondingangles)L7=60∘

In the given figure, l || m and p is the transversal. If $∠ 1$ and $∠ 2$ are in the ratio 1 : 2, find the measure of the angle $7$ in degrees