Question

In triangle ABC the measure of angle B is two - third of the measure of angle A. The measure of angle C is 20 degree more than the measure of angle A. Find the measure of the three angles

Answer 1

Answer:

$\angle A=60^{\circ}$, $\angle B=40^{\circ}$, and $\angle C=80^{\circ}$

Explanation:  

Given: In triangle ABC the measure of angle B is two - third of the measure of angle A. The measure of angle C is 20 degree more than the measure of angle A

To find: Measure of the three angles

Solution:

Let the measure of angle A be $x^{\circ}$. So, measure of angle B is $\frac{2}{3}x^{\circ}$, and measure of angle C is $(x+20)^{\circ}$

Now, we know that

$\angle A+\angle B+\angle C=180^{\circ}$  [Angle sum property]

$x+\frac{2}{3}x+x+20=180^{\circ}$

$2x+\frac{2}{3}x+20=180^{\circ}$

$2x+\frac{2}{3}x=160^{\circ}$

$8x=480^{\circ}$

$x=60^{\circ}$

So, $\angle A=60^{\circ}$, $\angle B=\frac{2}{3}\times60^{\circ}=40^{\circ}$, and $\angle C=(60+20)^{\circ}=80^{\circ}$

Learn more:

Find the measure of each angle of the triangle

https://brainly.in/question/5759916

Answer 2

Answer:

Let the measure of angle A be x^{\circ}x∘ . So, measure of angle B is \frac{2}{3}x^{\circ}32x∘ , and measure of angle C is (x+20)^{\circ}(x+20)∘

Now, we know that

\angle A+\angle B+\angle C=180^{\circ}∠A+∠B+∠C=180∘  [Angle sum property]

x+\frac{2}{3}x+x+20=180^{\circ}x+32x+x+20=180∘

2x+\frac{2}{3}x+20=180^{\circ}2x+32x+20=180∘

2x+\frac{2}{3}x=160^{\circ}2x+32x=160∘

8x=480^{\circ}8x=480∘

x=60^{\circ}x=60∘

So, \angle A=60^{\circ}∠A=60∘ , \angle B=\frac{2}{3}\times60^{\circ}=40^{\circ}∠B=32×60∘=40∘ , and \angle C=(60+20)^{\circ}=80^{\circ}∠C=(60+20)∘=80∘