Question
In $\triangle \text{ABC},$ if $3\angle \text{A}=4\angle \text{B}=6\angle \text{C},$ calculate the angles.
✓
Answer
calculate the angles. calculate the angles $3\angle \text{A}=6\angle \text{C}$
$\angle \text{A}=2\angle \text{C} ...(\text{i})$ We also know that for the same triangle, $4\angle \text{B}=6\angle \text{C}$
$\angle \text{B}=\frac{6}{4}\angle \text{C}...(\text{ii})$
We know that the sum of all three angles of a triangle is $180^\circ$
Therefore, we can say that: $\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$
$($Angles of $\triangle \text{ABC})...(\text{iii})$ On putting the values of $\angle \text{A}$ and $\angle \text{B}$ in equation $(iii),$
we get: $2\angle \text{C}+\Big(\frac{6}{4}\Big)\angle \text{C}+\angle \text{C}=180^\circ$
$\Big(\frac{18}{4}\Big)\angle \text{C}=180^\circ$
$\angle \text{C}=40^\circ$ From equation $(i),$
we have: $\angle \text{A}=2\angle \text{C}=2\times 40$
$\angle \text{A}=80^\circ$ From equation $(ii),$
we have: $\angle \text{B}=\Big(\frac{6}{4}\Big)\angle \text{C}=\Big(\frac{6}{4}\Big)\times40^\circ$
$\angle \text{B}=60^\circ$
$\angle \text{A}=80^\circ, \angle \text{B}=60^\circ, \angle \text{C}=40^\circ$ Therefore, the three angles of the given triangle are $80^\circ , 60^\circ $, and $40^\circ $
$\angle \text{A}=2\angle \text{C} ...(\text{i})$ We also know that for the same triangle, $4\angle \text{B}=6\angle \text{C}$
$\angle \text{B}=\frac{6}{4}\angle \text{C}...(\text{ii})$
We know that the sum of all three angles of a triangle is $180^\circ$
Therefore, we can say that: $\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$
$($Angles of $\triangle \text{ABC})...(\text{iii})$ On putting the values of $\angle \text{A}$ and $\angle \text{B}$ in equation $(iii),$
we get: $2\angle \text{C}+\Big(\frac{6}{4}\Big)\angle \text{C}+\angle \text{C}=180^\circ$
$\Big(\frac{18}{4}\Big)\angle \text{C}=180^\circ$
$\angle \text{C}=40^\circ$ From equation $(i),$
we have: $\angle \text{A}=2\angle \text{C}=2\times 40$
$\angle \text{A}=80^\circ$ From equation $(ii),$
we have: $\angle \text{B}=\Big(\frac{6}{4}\Big)\angle \text{C}=\Big(\frac{6}{4}\Big)\times40^\circ$
$\angle \text{B}=60^\circ$
$\angle \text{A}=80^\circ, \angle \text{B}=60^\circ, \angle \text{C}=40^\circ$ Therefore, the three angles of the given triangle are $80^\circ , 60^\circ $, and $40^\circ $
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