Question

22. In the given figure, l1 is parallel to l2 then find the value of x

110°
70°
100°
80'​

Answer 1

Answer:

Let the Triangle A,B,C=180

=35+45+x=180

=x=180-80

x=100

Let vertical opposite side=100

Angle x+100=180

=180-100

=80

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Answer 2

Step-by-step explanation:

Given: (from the figure below)

$I_{1} //I_{2}$

∠$OBC=45^{o}$

∠$BCO=35^{o}$

To Find: Value of $x$

Calculation:

From Δ$OBC$, sum of all the angles in a triangle is equal to $180^{o}$.

∠$O+$∠$B+$∠$C=180^{o}$

∠$O=180^{o} -(45^{o} +35^{o} )$

∠$O=100^{o}$ i.e. ∠$COB=100^{o}$

Since $I_{1}$ is parallel to $I_{2}$, $I_{1} //I_{2}$ then from corresponding angle, ∠$PAB=$∠$QOB$$=x$

From straight line angle pair, ∠$QOB+$∠$COB=180^{o}$

So, ∠$QOB=80^{o}$

Therefore, ∠$PAB=x=80^{o}$

Answer:

∠$PAB=x=80^{o}$

Value of $x$ is $80^{o}$.