
With usual notations, if in a triangle ABC,$\dfrac{{b + c}}{{11}} = \dfrac{{c + a}}{{12}} = \dfrac{{a + b}}{{13}}$ then $\cos A:\cos B:\cos C$is equal to
A) $\left( a \right){\text{ 7:19:25}}$
B) $\left( b \right){\text{ 19:7:25}}$
C) $\left( c \right){\text{ 12:14:20}}$
D) $\left( d \right){\text{ 19:25:20}}$
Answer
576.3k+ views
Hint:
For solving this type of question, we will assume the ratio be any arbitrary constant and from there we will calculate the sides of a triangle. And after that putting the values in the ratios we have to calculate, we will get the required ratios.
Complete step by step solution:
Let us assume each ratio be$k$, then we have the equation written as
$ \Rightarrow b + c = 11k,{\text{ }}c + a = 12k,{\text{ }}a + b = 13k$
Now from these three equations, on adding all the three equations, we will get
$ \Rightarrow 2\left( {a + b + c} \right) = 36k$
Taking the $2$to the right side, we get
$ \Rightarrow a + b + c = 18k$
Now using the above equation, after calculating the values we get
$ \Rightarrow a = 7k,{\text{ b = 6k, c = 5k}}$
So we have all the values of the term.
Now we will calculate the values of $\cos A:\cos B:\cos C$
So the values $\cos A$will be equals to
$ \Rightarrow \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$
On putting the values, we get
$ \Rightarrow \dfrac{{36{k^2} + 25{k^2} - 49{k^2}}}{{60{k^2}}}$
And on calculating, we get
$ \Rightarrow \dfrac{{12{k^2}}}{{60{k^2}}}$
And on doing the lowest common factor, we get
$ \Rightarrow \cos A = \dfrac{1}{5}$
So the values $\cos B$will be equals to
$ \Rightarrow \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}$
On putting the values, we get
$ \Rightarrow \dfrac{{49{k^2} + 25{k^2} - 36{k^2}}}{{70{k^2}}}$
And on calculating, we get
$ \Rightarrow \dfrac{{38{k^2}}}{{70{k^2}}}$
And on doing the lowest common factor, we get
$ \Rightarrow \cos B = \dfrac{{19}}{{35}}$
So the values $\cos C$will be equals to
$ \Rightarrow \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$
On putting the values, we get
$ \Rightarrow \dfrac{{49{k^2} + 36{k^2} - 25{k^2}}}{{84{k^2}}}$
And on calculating, we get
$ \Rightarrow \dfrac{{60{k^2}}}{{84{k^2}}}$
And on doing the lowest common factor, we get
$ \Rightarrow \cos A = \dfrac{5}{7}$
Therefore, the ratios $\cos A:\cos B:\cos C$will be
$ \Rightarrow \dfrac{1}{5}:\dfrac{{19}}{{35}}:\dfrac{5}{7}$
So it can also be written as
$ \Rightarrow 7:19:25$
Hence, the option $\left( a \right)$will be correct.
Note:
For solving this type of question, we should always assume the ratios be any constant and then go for the solution of it. In this type of question if our linear equations are correct then we can easily find out the ratios. And in this way, we can easily solve it.
For solving this type of question, we will assume the ratio be any arbitrary constant and from there we will calculate the sides of a triangle. And after that putting the values in the ratios we have to calculate, we will get the required ratios.
Complete step by step solution:
Let us assume each ratio be$k$, then we have the equation written as
$ \Rightarrow b + c = 11k,{\text{ }}c + a = 12k,{\text{ }}a + b = 13k$
Now from these three equations, on adding all the three equations, we will get
$ \Rightarrow 2\left( {a + b + c} \right) = 36k$
Taking the $2$to the right side, we get
$ \Rightarrow a + b + c = 18k$
Now using the above equation, after calculating the values we get
$ \Rightarrow a = 7k,{\text{ b = 6k, c = 5k}}$
So we have all the values of the term.
Now we will calculate the values of $\cos A:\cos B:\cos C$
So the values $\cos A$will be equals to
$ \Rightarrow \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$
On putting the values, we get
$ \Rightarrow \dfrac{{36{k^2} + 25{k^2} - 49{k^2}}}{{60{k^2}}}$
And on calculating, we get
$ \Rightarrow \dfrac{{12{k^2}}}{{60{k^2}}}$
And on doing the lowest common factor, we get
$ \Rightarrow \cos A = \dfrac{1}{5}$
So the values $\cos B$will be equals to
$ \Rightarrow \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}$
On putting the values, we get
$ \Rightarrow \dfrac{{49{k^2} + 25{k^2} - 36{k^2}}}{{70{k^2}}}$
And on calculating, we get
$ \Rightarrow \dfrac{{38{k^2}}}{{70{k^2}}}$
And on doing the lowest common factor, we get
$ \Rightarrow \cos B = \dfrac{{19}}{{35}}$
So the values $\cos C$will be equals to
$ \Rightarrow \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$
On putting the values, we get
$ \Rightarrow \dfrac{{49{k^2} + 36{k^2} - 25{k^2}}}{{84{k^2}}}$
And on calculating, we get
$ \Rightarrow \dfrac{{60{k^2}}}{{84{k^2}}}$
And on doing the lowest common factor, we get
$ \Rightarrow \cos A = \dfrac{5}{7}$
Therefore, the ratios $\cos A:\cos B:\cos C$will be
$ \Rightarrow \dfrac{1}{5}:\dfrac{{19}}{{35}}:\dfrac{5}{7}$
So it can also be written as
$ \Rightarrow 7:19:25$
Hence, the option $\left( a \right)$will be correct.
Note:
For solving this type of question, we should always assume the ratios be any constant and then go for the solution of it. In this type of question if our linear equations are correct then we can easily find out the ratios. And in this way, we can easily solve it.
Recently Updated Pages
Write a brief account of abscisic acid class 11 biology CBSE

Photolysis of water involves a Excitement of water class 11 biology CBSE

Both wind and water pollinated flowers are not very class 11 biology CBSE

Which among the following has specialized tissue for class 11 biology CBSE

Give one point of difference between the notochord class 11 biology CBSE

What are the factors that are essential for photos class 11 biology CBSE

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

State the laws of reflection of light

10 examples of friction in our daily life

Who is known as the father of chemistry class 11 chemistry CBSE

