Question

If in a triangle ABC, (a + b + c) (b + c – a) = k. bc, then
A. k < 0
B. k > 0
C. 0 < k < 4
D. k < 4

Answer 1

Hint: To solve this question, we will use the law of cosines which is used to find the angles of a triangle if we are given the sides of a triangle The law of cosine is $\cos {\text{A = }}\dfrac{{{{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{ - }}{{\text{a}}^2}}}{{{\text{2bc}}}}$.

Complete step-by-step answer:
Now, we are given (a + b + c) (b + c -a) = k. bc. We can also write it as,
(b + c + a) (b + c – a) = k. bc
Now, we can see that in the above equation the left – hand term is of the type (x + y) (x – y).
So, we will use the property (x + y) (x – y) = ${{\text{x}}^2}{\text{ - }}{{\text{y}}^2}$.so, we have
${{\text{(b + c)}}^2}{\text{ - }}{{\text{a}}^2}$ = k. bc
Now, we will use the identity ${({\text{a + b)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + 2ab}}$ to open the square in the above term. So, applying property, we get
${{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{ + 2bc - }}{{\text{a}}^2}$ = k. bc
${{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{ - }}{{\text{a}}^2}$ = k. bc – 2bc
Taking bc common in R. H. S, we get
${{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{ - }}{{\text{a}}^2}$ = (k – 2) bc
It can also be written as, $\dfrac{{{{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{ - }}{{\text{a}}^2}}}{{{\text{bc}}}}$ = k – 2
Now, dividing both sides by 2, we get
$\dfrac{{{{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{ - }}{{\text{a}}^2}}}{{{\text{2bc}}}}$ = $\dfrac{{{\text{k - 2}}}}{2}$
Now, by law of cosine, $\cos {\text{A = }}\dfrac{{{{\text{b}}^2}{\text{ + }}{{\text{c}}^2}{\text{ - }}{{\text{a}}^2}}}{{{\text{2bc}}}}$. We can find any angle with the help of the law of cosine from the figure drawn below.

Therefore, cos A = $\dfrac{{{\text{k - 2}}}}{2}$ … (1)
Now, cos A has a maximum value of 1 and minimum value of -1. So, we have
${\text{ - 1}} \leqslant {\text{ cosA }} \leqslant {\text{1}}$
Putting value of cos A from equation (1), we have
${\text{ - 1}} \leqslant {\text{ }}\dfrac{{{\text{k - 2}}}}{2} \leqslant {\text{1}}$
Multiplying the above term by 2, we get
${\text{ - 2}} \leqslant {\text{ k - 2}} \leqslant 2$
Adding 2 in the above term, we get
${\text{ - 2 + }}{\text{ 2}} \leqslant {\text{ k - 2 + }}{\text{ 2}} \leqslant 2{\text{ + 2}}$
$0 \leqslant {\text{ k }} \leqslant 4$
So, 0 < k < 4
So, option (3) is the correct answer.

Note: When we come up with such types of question, we will first simplify the given expression by using the properties of (x + y)(x – y) = ${{\text{x}}^2}{\text{ - }}{{\text{y}}^2}$ and ${({\text{a + b)}}^2}{\text{ = }}{{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ + 2ab}}$. After it, we will use the law of cosine to find the value of the variable asked in the question.