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Find the measure of $\angle BCD$.
"
Given:
$\angle ABD=70^o, \angle ADB=30^o$.
To do:
We have to find $\angle BCD$.
Solution:
In $\vartriangle ADB$,
$\angle ABD+\angle ADB+\angle BAD=180^o$
$70^o+30^o+\angle BAD=180^o$
$\angle BAD=(180-100)^o=80^o$
We know that,
The sum of angles in opposite segments of a circle is $180^o$.
Therefore,
$\angle BAD+\angle BCD=180^o$
$80^o+\angle BCD=180^o$
$\angle BCD=(180-80)^o$
$\angle BCD=100^o$
The measure of $\angle BCD$ is $100^o$.
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