# If in a triangle ABC $\cos A\cos B + \sin A\sin B\sin C = 1$, show that $a:b:c = 1:1:\sqrt 2 $.

Last updated date: 19th Mar 2023

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Answer

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Hint: In this question use the concept that the value of sin is always less than or equal to 1, later on apply that in a triangle ABC the ratio of its sides a, b and c are equal to the ratio of sin of the angles A, B and C so use these concepts to reach the solution of the question.

Complete step-by-step answer:

Given equation is

$\cos A\cos B + \sin A\sin B\sin C = 1$

Proof –

From the above equation simplify for sin C we have,

$ \Rightarrow \sin A\sin B\sin C = 1 - \cos A\cos B$

$ \Rightarrow \sin C = \dfrac{{1 - \cos A\cos B}}{{\sin A\sin B}}$ ……………………. (1)

Now as we know that $\sin C \leqslant 1$

Therefore from above equation we have,

$ \Rightarrow \dfrac{{1 - \cos A\cos B}}{{\sin A\sin B}} \leqslant 1$

Now simplify this equation we have,

$ \Rightarrow 1 - \cos A\cos B \leqslant \sin A\sin B$

$ \Rightarrow 1 \leqslant \sin A\sin B + \cos A\cos B$

Now as we know $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ so use this property in above equation we have,

$ \Rightarrow 1 \leqslant \cos \left( {A - B} \right)$ or $ \Rightarrow \cos \left( {A - B} \right) \geqslant 1$

Now above inequality satisfies for $\left[ {\cos \left( {A - B} \right) = 1} \right]$, ($\cos \left( {A - B} \right) > 1$ is never possible).

$ \Rightarrow \cos \left( {A - B} \right) = 1 = \cos 0$

$ \Rightarrow A - B = 0$ or $A = B$ ……………………… (2)

Now from equation (1) we have,

$ \Rightarrow \sin C = \dfrac{{1 - \cos A\cos A}}{{\sin A\sin A}} = \dfrac{{1 - {{\cos }^2}A}}{{{{\sin }^2}A}}$

Now as we know that $\left( {1 - {{\cos }^2}A = {{\sin }^2}A} \right)$

$ \Rightarrow \sin C = 1 = \sin {90^0}$

So on comparing,

$ \Rightarrow C = {90^0}$…………………. (3)

Now as we know in triangle the sum of all angles is 180 degree.

$ \Rightarrow A + B + C = {180^0}$

Ow from equation (2) and (3) we have,

$ \Rightarrow A + A + {90^0} = {180^0}$

$ \Rightarrow 2A = {90^0}$

$ \Rightarrow A = {45^0} = B$

Now as we know in a triangle;

$a:b:c = \sin A:\sin B:\sin C$

Now substitute the value of A, B and C in above equation we have,

$ \Rightarrow a:b:c = \sin {45^0}:\sin {45^0}:\sin {90^0}$

Now substitute the values of sine angles we have,

$ \Rightarrow a:b:c = \dfrac{1}{{\sqrt 2 }}:\dfrac{1}{{\sqrt 2 }}:1$

Multiply by $\sqrt 2 $ we have,

$ \Rightarrow a:b:c = 1:1:\sqrt 2 $

Hence Proved.

Note: In such types of questions the key concept we have to remember is that always recall the condition of sin which is stated above and the sum of all the angles in a triangle is equal to 180 degree so using these properties simplify the equation as above and calculate all the values of the angles then apply the property that in a triangle ABC the ratio of its sides a, b and c are equal to the ratio of sin of the angles A, B and C so use this and calculate the ratio which is the required answer.

Complete step-by-step answer:

Given equation is

$\cos A\cos B + \sin A\sin B\sin C = 1$

Proof –

From the above equation simplify for sin C we have,

$ \Rightarrow \sin A\sin B\sin C = 1 - \cos A\cos B$

$ \Rightarrow \sin C = \dfrac{{1 - \cos A\cos B}}{{\sin A\sin B}}$ ……………………. (1)

Now as we know that $\sin C \leqslant 1$

Therefore from above equation we have,

$ \Rightarrow \dfrac{{1 - \cos A\cos B}}{{\sin A\sin B}} \leqslant 1$

Now simplify this equation we have,

$ \Rightarrow 1 - \cos A\cos B \leqslant \sin A\sin B$

$ \Rightarrow 1 \leqslant \sin A\sin B + \cos A\cos B$

Now as we know $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ so use this property in above equation we have,

$ \Rightarrow 1 \leqslant \cos \left( {A - B} \right)$ or $ \Rightarrow \cos \left( {A - B} \right) \geqslant 1$

Now above inequality satisfies for $\left[ {\cos \left( {A - B} \right) = 1} \right]$, ($\cos \left( {A - B} \right) > 1$ is never possible).

$ \Rightarrow \cos \left( {A - B} \right) = 1 = \cos 0$

$ \Rightarrow A - B = 0$ or $A = B$ ……………………… (2)

Now from equation (1) we have,

$ \Rightarrow \sin C = \dfrac{{1 - \cos A\cos A}}{{\sin A\sin A}} = \dfrac{{1 - {{\cos }^2}A}}{{{{\sin }^2}A}}$

Now as we know that $\left( {1 - {{\cos }^2}A = {{\sin }^2}A} \right)$

$ \Rightarrow \sin C = 1 = \sin {90^0}$

So on comparing,

$ \Rightarrow C = {90^0}$…………………. (3)

Now as we know in triangle the sum of all angles is 180 degree.

$ \Rightarrow A + B + C = {180^0}$

Ow from equation (2) and (3) we have,

$ \Rightarrow A + A + {90^0} = {180^0}$

$ \Rightarrow 2A = {90^0}$

$ \Rightarrow A = {45^0} = B$

Now as we know in a triangle;

$a:b:c = \sin A:\sin B:\sin C$

Now substitute the value of A, B and C in above equation we have,

$ \Rightarrow a:b:c = \sin {45^0}:\sin {45^0}:\sin {90^0}$

Now substitute the values of sine angles we have,

$ \Rightarrow a:b:c = \dfrac{1}{{\sqrt 2 }}:\dfrac{1}{{\sqrt 2 }}:1$

Multiply by $\sqrt 2 $ we have,

$ \Rightarrow a:b:c = 1:1:\sqrt 2 $

Hence Proved.

Note: In such types of questions the key concept we have to remember is that always recall the condition of sin which is stated above and the sum of all the angles in a triangle is equal to 180 degree so using these properties simplify the equation as above and calculate all the values of the angles then apply the property that in a triangle ABC the ratio of its sides a, b and c are equal to the ratio of sin of the angles A, B and C so use this and calculate the ratio which is the required answer.

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