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

Last updated date: 06th Aug 2024
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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.

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.