
In $\Delta $ ABC, $3\angle A = 4\angle B = 6\angle C.$ The angles of triangle are:
A. $A = {60^0},B = {80^0}{\text{ and }}C = {40^0}$
B. $A = {60^0},B = {65^0}{\text{ and }}C = {55^0}$
C. $A = {70^0},B = {30^0}{\text{ and }}C = {80^0}$
D. $A = {80^0},B = {60^0}{\text{ and }}C = {40^0}$
Answer
617.1k+ views
Hint: In order to solve such a type of question, we will use the sum of angle property of a triangle which states that the sum of angles of any triangle must be 180 degrees. We will proceed by converting the given conditions in terms of one variable and with the use angle sum property solved further.
Complete step-by-step answer:
Given that $3\angle A = 4\angle B = 6\angle C.$
Now, convert both A and B in terms of C
So, $\angle A = 2\angle C{\text{ and }}\angle B = \dfrac{3}{2}\angle C........(1)$
As we know that sum of angles of triangle is ${180^{^0}}$
$\therefore \angle A + \angle B + \angle C = {180^0}$
Put the values of $\angle A = 2\angle C{\text{ and }}\angle B = \dfrac{3}{2}\angle C$ in above equation, we get
\[
\Rightarrow 2\angle C + \dfrac{3}{2}\angle C + \angle C = {180^0} \\
\Rightarrow \angle C\left( {2 + \dfrac{3}{2} + 1} \right) = {180^0} \\
\Rightarrow \angle C\left( {\dfrac{{4 + 3 + 2}}{2}} \right) = {180^0} \\
\Rightarrow \angle C = {180^0} \times \dfrac{2}{9} \\
\Rightarrow \angle C = {40^0} \\
\]
By substituting \[\angle C = {40^0}\] in equation (1), we get
\[
\Rightarrow \angle A = 2\angle C = 2 \times {40^0} = {80^0} \\
\Rightarrow \angle B = \dfrac{3}{2}\angle C = \dfrac{3}{2} \times {40^0} = {60^0} \\
\]
Hence, the value of \[\angle A = {80^0},\angle B = {60^0}{\text{ and }}\angle C = {40^0}\] and the correct answer id option “D”.
Note: In order to solve this type of problem, first we need to remember all the properties of the triangle like we use angle sum property in the above question. Also the sum properties of a triangle are the sum of the two sides of a triangle is always greater than the length of the third side .A triangle has three sides, three vertices and three angles. Second, you must be good at basic algebra.
Complete step-by-step answer:
Given that $3\angle A = 4\angle B = 6\angle C.$
Now, convert both A and B in terms of C
So, $\angle A = 2\angle C{\text{ and }}\angle B = \dfrac{3}{2}\angle C........(1)$
As we know that sum of angles of triangle is ${180^{^0}}$
$\therefore \angle A + \angle B + \angle C = {180^0}$
Put the values of $\angle A = 2\angle C{\text{ and }}\angle B = \dfrac{3}{2}\angle C$ in above equation, we get
\[
\Rightarrow 2\angle C + \dfrac{3}{2}\angle C + \angle C = {180^0} \\
\Rightarrow \angle C\left( {2 + \dfrac{3}{2} + 1} \right) = {180^0} \\
\Rightarrow \angle C\left( {\dfrac{{4 + 3 + 2}}{2}} \right) = {180^0} \\
\Rightarrow \angle C = {180^0} \times \dfrac{2}{9} \\
\Rightarrow \angle C = {40^0} \\
\]
By substituting \[\angle C = {40^0}\] in equation (1), we get
\[
\Rightarrow \angle A = 2\angle C = 2 \times {40^0} = {80^0} \\
\Rightarrow \angle B = \dfrac{3}{2}\angle C = \dfrac{3}{2} \times {40^0} = {60^0} \\
\]
Hence, the value of \[\angle A = {80^0},\angle B = {60^0}{\text{ and }}\angle C = {40^0}\] and the correct answer id option “D”.
Note: In order to solve this type of problem, first we need to remember all the properties of the triangle like we use angle sum property in the above question. Also the sum properties of a triangle are the sum of the two sides of a triangle is always greater than the length of the third side .A triangle has three sides, three vertices and three angles. Second, you must be good at basic algebra.
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