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# In a triangle $ABC$, if $\cot A:\cot B:\cot C = 30:19:6$, then the sides $a, b, c$ are in:$\left( {\text{A}} \right){\text{ }}$ In A.P.$\left( {\text{B}} \right){\text{ }}$ In G.P.$\left( {\text{C}} \right){\text{ }}$ In H.P.$\left( {\text{D}} \right){\text{ }}$ None of these

Last updated date: 09th Aug 2024
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Hint: To solve this question we need to use the formula and then split it to find the value of $a,b,c$.
Then we have to apply a condition to check if the solution is correct are not.
Next we must observe the series which we get and mention the series which it is representing.
Finally we get the required answer.

Formula used: $\cot A = \dfrac{{b_{}^2 + c_{}^2 - a_{}^2}}{{4\Delta }}$,$\cot B = \dfrac{{c_{}^2 + a_{}^2 - b_{}^2}}{{4\Delta }}$ and $\cot C = \dfrac{{b_{}^2 + a_{}^2 - c_{}^2}}{{4\Delta }}$.

It is given in the question $\cot A:\cot B:\cot C = 30:19:6$,
Now by applying the above formula of $\cot A,\cot B,\cot C$ in the given equation we get-
$\dfrac{{b_{}^2 + c_{}^2 - a_{}^2}}{{4\Delta }}:\dfrac{{c_{}^2 + a_{}^2 - b_{}^2}}{{4\Delta }}:\dfrac{{b_{}^2 + a_{}^2 - c_{}^2}}{{4\Delta }} = 30:19:6$
Now by breaking the above equation we get-
$\Rightarrow$$\dfrac{{b_{}^2 + {c^2} - a_{}^2}}{{4\Delta }} = 30....\left( 1 \right) \Rightarrow$$\dfrac{{c_{}^2 + a_{}^2 - b_{}^2}}{{4\Delta }} = 19....\left( 2 \right)$
$\Rightarrow$$\dfrac{{b_{}^2 + a_{}^2 - c_{}^2}}{{4\Delta }} = 6....\left( 3 \right) Now from equation \left( 1 \right) we get- \Rightarrow$$b_{}^2 + c_{}^2 - a_{}^2 = 30(4\Delta ) = 30k_{}^2....\left( 4 \right)$, where $k_{}^2$ is a constant
Now from equation $\left( 2 \right)$ we get-
$\Rightarrow$$c_{}^2 + a_{}^2 - b_{}^2 = 19(4\Delta ) = 19k_{}^2....\left( 5 \right), where k_{}^2 is a constant Now from equation \left( 3 \right) we get- \Rightarrow$$b_{}^2 + a_{}^2 - c_{}^2 = 6(4\Delta ) = 6k_{}^2....\left( 6 \right)$, where $k_{}^2$ is a constant
Now, by adding equation$\left( 4 \right)$ ,$\left( 5 \right)$ and $\left( 6 \right)$ we get-
$\Rightarrow$$b_{}^2 + c_{}^2 - a_{}^2 + c_{}^2 + a_{}^2 - b_{}^2 + b_{}^2 + a_{}^2 - c_{}^2 = 30k_{}^2 + 19k_{}^2 + 6k_{}^2 So we can write- \Rightarrow$$a_{}^2 + b_{}^2 + c_{}^2 = 55k_{}^2....\left( 7 \right)$
Now by subtracting equation $\left( 4 \right)$ from equation $\left( 7 \right)$ we get-
$a_{}^2 + b_{}^2 + c_{}^2 - b_{}^2 - c_{}^2 + a_{}^2 = 55k_{}^2 - 30k_{}^2$
On some simplification we get,
$\Rightarrow$$2a_{}^2 = 25k_{}^2 Now by division we get- \Rightarrow$$a_{}^2 = \dfrac{{25k_{}^2}}{2}$
By applying square root we get-
$\Rightarrow$$a = \sqrt {\dfrac{{25k_{}^2}}{2}} = \dfrac{{5k}}{{\sqrt 2 }} Again by subtracting equation \left( 5 \right) from equation \left( 7 \right) we get- \Rightarrow$$a_{}^2 + b_{}^2 + c_{}^2 - c_{}^2 - a_{}^2 + b_{}^2 = 55k_{}^2 - 19k_{}^2$
On some simplification we get,
$\Rightarrow$$2b_{}^2 = 36k_{}^2 Now by division we get- \Rightarrow$$b_{}^2 = \dfrac{{36k_{}^2}}{2}$
By doing square root we get-
$\Rightarrow$$b_{}^{} = \sqrt {\dfrac{{36k_{}^2}}{2}} = \dfrac{{6k}}{{\sqrt 2 }} Again by subtracting equation \left( 6 \right) from \left( 7 \right) we get- \Rightarrow$$a_{}^2 + b_{}^2 + c_{}^2 - b_{}^2 - a_{}^2 + c_{}^2 = 55k_{}^2 - 6k_{}^2$
On some simplification we get,
$\Rightarrow$$c_{}^2 = \dfrac{{49k_{}^2}}{2} Now by doing squaring both side we get- \Rightarrow$$c = \dfrac{{7k}}{{\sqrt 2 }}$
Therefore the value of $a = \dfrac{{5k}}{{\sqrt 2 }}$
The value of $b = \dfrac{{6k}}{{\sqrt 2 }}$
The value of $c = \dfrac{{7k}}{{\sqrt 2 }}$
From the above values, it can be said that the sides $a, b, c$ are in A.P as it satisfies the condition $2b = a + c$.

Option A is the correct answer.

Note: In order to check whether the progression is in A.P. or not we have to apply the formula $2b = a + c$ where the twice of the middle term will be equal to the addition of the first and last term and if the condition is satisfied then the numbers are in A.P.
Verification:
Substitute the value in the condition $2b = a + c$ and we get,
$2\left( {\dfrac{{6k}}{{\sqrt 2 }}} \right) = \dfrac{{5k}}{{\sqrt 2 }} + \dfrac{{7k}}{{\sqrt 2 }}$
On multiply the RHS and add the terms as LHS we get,
$\dfrac{{12k}}{{\sqrt 2 }} = \dfrac{{12k}}{{\sqrt 2 }}$
Hence satisfies the condition that the sides $a, b, c$are in A.P.