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