In the accomapnying diagram of ABC, BC is extended to D, m<A = 20*, m<ACD = 70*, and m<B = x*.
* = degress
_ = Arrow over
Anyone mind explaining?
Value of X.
Hello, bon22!
$\displaystyle \angle ACD$ is an exterior angle of $\displaystyle \Delta ABC$In the accomapnying diagram of $\displaystyle \Delta ABC$,
$\displaystyle BC\text{ is extended to }D,\;\;\angle A = 20^o,\;\;\angle ACD = 70^o$
Find $\displaystyle \angle B$Code:A o * * *20°* * * * * * * 70° B o * * * o * * o D C
Hence, it equals the sum of the two non-adjacent angles, $\displaystyle \angle A\text{ and }\angle B.$
Therefore: .$\displaystyle \angle B \:=\:50^o$