Question

If in a triangle ABC,$\text{cosAcosB + sinAsinBsinC = 1}$, then find the ratio a:b:c
A. $1:1:\sqrt{2}$
B. $1:1:\sqrt{3}$
C. \[1:\sqrt{2}:1\]
D. $1:\sqrt{3}:1$

Answer 1

Hint: Reduce the given relation to find relations between the angles of the triangle. Thus, find the nature of the triangle from these relations. Then find the ratios of the sides of the triangles. The equation can be solved by using the identity $sin^2 x + cos^2 x =1$ and the property that if the terms on LHS are non-negative and RHS is 0, then all terms are 0.
Lastly, by simplifying the given equation by the help of trigonometric formula, we will get our required answer.

Complete step-by-step answer:
Given, $\text{cosAcosB + sinAsinBsinC = 1}$
Reducing this given equation in the following way,
\[\begin{align}
  & \text{cosAcosB + sinAsinBsinC = 1} \\
 & \Rightarrow \text{ 2cosAcosB + 2sinAsinBsinC = 2 }\left( \text{--Multiplying 2 on both sides} \right) \\
 & \Rightarrow \text{ 2cosAcosB + 2sinAsinBsinC = }\left( \text{si}{{\text{n}}^{2}}\text{A + co}{{\text{s}}^{2}}\text{A} \right)\text{ + }\left( \text{si}{{\text{n}}^{2}}\text{B + co}{{\text{s}}^{2}}\text{B} \right) \\
 & \Rightarrow \text{ }\left( \text{co}{{\text{s}}^{2}}\text{A + co}{{\text{s}}^{2}}\text{B }-\text{ 2cosAcosB} \right)\text{ + }\left( \text{si}{{\text{n}}^{2}}\text{A + si}{{\text{n}}^{2}}\text{B }-\text{ 2sinAsinB} \right)\text{ + 2sinAsinB }-\text{ 2sinAsinBsinC = 0} \\
 & \Rightarrow \text{ }{{\left( \text{cosA }-\text{ cosB} \right)}^{2}}\text{ + }{{\left( \text{sinA }-\text{ sinB} \right)}^{2}}\text{ + 2sinAsinB}\left( 1\text{ }-\text{ sinC} \right)\text{ = 0 }......\text{(i)} \\
\end{align}\]Notice that all the quantities in the left hand side of the equation (i) are non-negative quantities. The first two quantities are square quantities, so naturally they are non-negative. Now, since A, B and C are sides of a triangle, 0 < A, B, C < $\pi $, thus the sine of all the angles are greater than 0.
Thus, for the sum of the non-negative quantities to be equal to 0, each of the quantities is equal to zero.
Therefore,
$\begin{align}
  & \text{sinA = sinB }....\left( \text{ii} \right) \\
 & \text{cosA = cosB }....\left( \text{iii} \right) \\
 & \text{1 }-\text{ sinC = 0 }\left( \because \sin \text{A and sinB can }\!\!'\!\!\text{ t be equal to zero, otherwise it won }\!\!'\!\!\text{ t remain a triangle} \right) \\
 & \Rightarrow \text{ sinC = 1 }....\text{(iv)} \\
\end{align}$
From (ii) and (iii), it can be said that, A = B
From (iv), it can be said that $\text{C = }\dfrac{\pi }{2}$
Thus, triangle ABC is a right-angled isosceles triangle where angles A and B are equal and angle C is right-angled.
Thus, the sides ‘a’ and ‘b’ are equal correspondingly, since ABC is isosceles and side ‘c’ is the hypotenuse.
If the length of each of ‘a’ and ‘b’ are x units, then according to Pythagoras.’ Theorem , which states that,
$\begin{align}
  & {{\text{a}}^{2}}\text{ + }{{\text{b}}^{2}}\text{ = }{{\text{c}}^{2}} \\
 & \Rightarrow \text{ }{{x}^{2}}\text{ + }{{x}^{2}}\text{ = }{{\text{c}}^{2}} \\
 & \therefore \text{ c = x}\sqrt{2} \\
\end{align}$
Thus, the ratio a:b:c
$\begin{align}
  & =\text{ }x:x:x\sqrt{2} \\
 & =\text{ 1:1:}\sqrt{2} \\
\end{align}$
Hence, the correct answer is option A.

Note: While solving the question, students should remember that $sin^2 x + cos^2 x =1$. It is a very commonly used property in trigonometry. Also, it must be kept in mind that for the sum of the non-negative quantities to be equal to 0, each of the quantities is equal to zero. Without this property, the question can’t be solved.