Question

Question:—
In $\Delta$ABC, if 3$\angle$A = 4$\angle$B = 6$\angle$C. Then, calculate the value of $\angle$A, $\angle$B, $\angle$C.
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Answer 1

Step-by-step explanation:

A=x/3 B=x/4 c=x/6

A+B+C=180

x/3+x/4+x/6=180

9x/12=180

x=240

a=80 B=60 C=40

Answer 2

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Answer:

Given:

• $\sf{3 \angle A = 4 \angle B = 6 \angle C}$

To find :

• $\sf{\angle A , \angle B , \angle C}$

$\mathfrak{\underline{Solution:-}}$

We are given the ratio of all angles. So we can write all angles as the fraction of one angle. For this we take one part lf first two angles then the other two.

$3 \angle A = 4 \angle B = 6 \angle C$

✦Now first taking angles A and B :

= 3$\angle$A = 4

= $\angle$A = $\dfrac{4}{3}$

✦Now taking angles B and C :

= 4 $\angle$B = 6

= $\angle$B = $\dfrac{{6\!\!\!\!／}^{{}^{3}}}{{4\!\!\!\!／}^{{}^{2}}}$

= $\angle$B = $\dfrac{3}{2} \angle$C

✦Now putting value of B from (ii) in (i):

= $\angle$A = $\dfrac{{4\!\!\!\!／}^{{}^{2}}}{3\!\!\!\!／} \times \dfrac{3\!\!\!\!／}{2\!\!\!\!／} \angle C$

= $\angle$A = 2

☘Using the angle sum property:

$\boxed{\sf{\angle A + \angle B + \angle C = 180^0}}$

➣Plugging in all the values:

= $\angle$A + $\dfrac{3}{4} \angle$A + $\dfrac{1}{2} \angle$A = 180⁰

= $\dfrac{4 \angle A + 3 \angle A + 2 \angle A}{4} = 180^0$

= 9$\angle$ A = 180⁰ × 4

= $\angle$A = $\dfrac{180^0 \times 4}{9}$

= $\angle$A = 80⁰

➣Now using (i) , (ii) , (iii) , We find the other angles as :

= $\angle$B = $\dfrac{3}{4} \times 80 = 60$

=$\angle$C = $\dfrac{1}{2} \times 80 = 40$

Hope it helps❤❤❤