Question:
In the given figure, $I \| m$ and a transversal $t$ cuts them. If $\angle 7=80^{\circ}$, find the measure of each of the remaining marked angles.
Solution:
In the given figure, ∠7 and ∠8 form a linear pair.
∴ ∠7 + ∠8 = 180º
⇒ 80º + ∠8 = 180º
⇒ ∠8 = 180º − 80º = 100º
Now,
∠6 = ∠8 = 100º (Vertically opposite angles)
∠5 = ∠7 = 80º (Vertically opposite angles)
It is given that, l || m and t is a transversal.
∴ ∠1 = ∠5 = 80º (Pair of corresponding angles)
∠2 = ∠6 = 100º (Pair of corresponding angles)
∠3 = ∠7 = 80º (Pair of corresponding angles)
∠4 = ∠8 = 100º (Pair of corresponding angles)
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