# 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

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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.

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.

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.

Last updated date: 21st Sep 2023

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