Question

15. In AABC if mZA is thirteen less than mzC and mZB is eleven less than four times m

Answer 1

Answer:

$m\angle A=21$°

$m\angle B=125$°

$m\angle C=34$°

Step-by-step explanation:

Let the measure of angle C be $x$°.

Given:

In triangle ΔABC,

$m\angle A$ is thirteen less than $m\angle C$ and $m\angle B$ is eleven less than four times $m\angle C$. This gives,

$m\angle A = x-13$

$m\angle B=4x-11$

Also, $m\angle C=x$

Now, for a triangle, the sum of all its interior angles is equal to 180°.

Therefore, $m\angle A + m\angle B + m\angle C=180$

Plug in all the values and solve for x. This gives,

$x-13+4x-11+x=180\\6x-24=180\\6x=180+24\\6x=204\\x=\frac{204}{6}=34$

Therefore, measure of angle C is 34°.

Measure of angle A is, $m\angle A=x-13=34-13=21$°.

Measure of angle B is, $m\angle B=4x-11=4(34)-11=136-11=125$°.