Question

. ZA= ZB+ 2C
ZA+ ZA = ZA+B+C =180°
-0. In a ABC, if 2 ZA = 3 ZB = 6 ZC, calculate ZA, ZB and ZC​

Answer 1

Answer:

C=120

∘

\angle B=40^{\circ}∠B=40

∘

\angle A=20^{\circ}∠A=20

∘

Step-by-step explanation:

In triangle ABC

\angle C=3\angle B=2(\angle A+\angle B)∠C=3∠B=2(∠A+∠B)

\angle A+\angle B=\frac{1}{2}\angle C∠A+∠B=

2

1

∠C

\angle B=\frac{1}{3}\angle C∠B=

3

1

∠C

\angle A+\angle B+\angle C=180^{\circ}∠A+∠B+∠C=180

∘

By using triangle angles sum property

Substitute the values then we get

\frac{1}{2}\angle C+\angle C=180

2

1

∠C+∠C=180

\frac{\angle C+2\angle C}{2}=180

2

∠C+2∠C

=180

\frac{3}{2}\angle C=180

2

3

∠C=180

\angle C=\frac{180\times 2}{3}=120^{\circ}∠C=

3

180×2

=120

∘

\angle C=120^{\circ}∠C=120

∘

Substitute the values then we get

\angle B=\frac{1}{3}\times 120=40^{\circ}∠B=

3

1

×120=40

∘

Substitute the values then we get

2(\angle A+40)=1202(∠A+40)=120

2\angle A+80=1202∠A+80=120

2\angle A=120-80=402∠A=120−80=40

\angle A=\frac{40}{2}=20^{\circ}∠A=

2

40

=20

∘

Answer 2

Answer:

refer to the attachment above

Step-by-step explanation:

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