In the given figure- lines AB and CD intersect point O-Find the value of angle a-

Question

Toppr Admin · Accepted Answer

Given-AB- and -CD- intersect each other at point -C-angle AOC-68-o-Here -AOB-160- is a lineHence-angle AOC-angle BOC -180-o- -Linear pair of angles-68-o -a-180-o-a-180-o-68-o-a-112-o-

In the given figure, lines $$AB$$ and $$CD$$ intersect at point $$O$$.
Find the value of $$\angle a$$.

Given
$$AB$$ and $$CD$$ intersect each other at point $$C$$
$$\angle AOC=68^o$$
Here $$AOB$$ is a line
Hence,
$$\angle AOC+\angle BOC =180^o$$ [Linear pair of angles]
$$68^o +a=180^o$$
$$a=180^o-68^o$$
$$a=112^o$$

In the given figure, lines $$AB$$ and $$CD$$ intersect at point $$O$$.Find the value of $$\angle a$$.

Given$$AB$$ and $$CD$$ intersect each other at point $$C$$$$\angle AOC=68^o$$Here $$AOB$$ is a lineHence,$$\angle AOC+\angle BOC =180^o$$ [Linear pair of angles]$$68^o +a=180^o$$$$a=180^o-68^o$$$$a=112^o$$

In the given figure, lines $$AB$$ and $$CD$$ intersect at point $$O$$.
Find the value of $$\angle a$$.

Given
$$AB$$ and $$CD$$ intersect each other at point $$C$$
$$\angle AOC=68^o$$
Here $$AOB$$ is a line
Hence,
$$\angle AOC+\angle BOC =180^o$$ [Linear pair of angles]
$$68^o +a=180^o$$
$$a=180^o-68^o$$
$$a=112^o$$