Question

If in a triangle ABC, $\cos A\cos B + \sin A\sin B\sin C = 1$, then find the value of $\dfrac{{a + b}}{c}$.

Answer 1

Hint: Find the relation between the angles A, B and C from the given equation to find the type of the triangle, then find the ratio involving the sides.

Complete step-by-step answer:
For triangle ABC, we solve to find the relation between the angles A, B and C.
It is given that,
$\cos A\cos B + \sin A\sin B\sin C = 1..........(1)$
We know that the value of sine is less than or equal to 1.
$\sin C \leqslant 1$
We, now, multiply sinA sinB on both sides of the inequality.
$\sin A\sin B\sin C \leqslant \sin A\sin B$
We, now, add the term cosA cosB to both sides of the inequality.
$\cos A\cos B + \sin A\sin B\sin C \leqslant \cos A\cos B + \sin A\sin B$
The left-hand side of the inequality is equal to 1 from equation (1).
$1 \leqslant \cos A\cos B + \sin A\sin B$
The right-hand side is the cosine of difference between the angles A and B.
$1 \leqslant \cos (A - B)$
We know that the value of cosine of any angle is less than or equal to one, then, we have:
$1 \leqslant \cos (A - B) \leqslant 1$
Hence, the value of cos(A – B) is 1.
$\cos (A - B) = 1$
We know that the value of cos0 is equal to one, then the angles A and B are equal.
$A = B..........(2)$
Substituting equation (2) in equation (1), we get:
${\cos ^2}A + {\sin ^2}A\sin C = 1$
We know that the value of ${\cos ^2}A + {\sin ^2}A$ is equal to one, then, we have the value of sinC equal to one.
$\sin C = 1$
$C = 90^\circ$
Hence, triangle ABC is isosceles right angles triangle with right angle at C.
Hence, the sides a and b are equal and c is the hypotenuse.
Using, Pythagoras theorem, we have:
${a^2} + {b^2} = {c^2}$
Since, sides a and b are equal, we have:
$2{a^2} = {c^2}$
Finding ratio of a and c, we have:
$\dfrac{{{a^2}}}{{{c^2}}} = \dfrac{1}{2}$
$\dfrac{a}{c} = \dfrac{1}{{\sqrt 2 }}........(3)$
We need to find the value of $\dfrac{{a + b}}{c}$, we know that a and b are equal, therefore, we have:
$\dfrac{{a + b}}{c} = \dfrac{{2a}}{c}$
Using equation (3), we get:
$\dfrac{{a + b}}{c} = \dfrac{2}{{\sqrt 2 }}$
$\dfrac{{a + b}}{c} = \sqrt 2$
Hence, the value is $\sqrt 2$.

Note: You know that the sum of two sides is always greater than the third side, so the answer should be greater than 1, this fact can be used for cross-checking.