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In a triangle $ABC$, $a = 4,b = 3,\angle A = 60^\circ $. Then $c$ is the root of which of the following equations?
A) ${c^2} - 3c - 7 = 0$
B) ${c^2} + 3c + 7 = 0$
C) ${c^2} - 3c + 7 = 0$
D) ${c^2} + 3c - 7 = 0$

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Last updated date: 20th Jun 2024
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Answer
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Hint: Here we are given two sides and the angle of a triangle. So we can use the cosine rule to find the equation. Substituting the given data and simplifying we get the required equation.

Formula used: The cosine rule states that the square of the length of any side of a triangle equals the sum of the squares of the length of the other sides minus twice the product multiplied by the cosine of their included angle.
In symbols we have,
${a^2} = {b^2} + {c^2} - 2bc\cos A$
Or it can be also written as $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$

Complete step-by-step solution:
Given that in the triangle $ABC$, $a = 4,b = 3,\angle A = 60^\circ $.
We need to find the equation with $c$ as its root.
Here by conventional notation, $a,b,c$ represents the sides opposite to vertices $A,B,C$ respectively.
So we are given two sides and the included angle of the triangle.
Therefore we can apply the cosine rule here.
The cosine rule states that the square of the length of any side of a triangle equals the sum of the squares of the length of the other sides minus twice the product multiplied by the cosine of their included angle.
In symbols we have,
${a^2} = {b^2} + {c^2} - 2bc\cos A$
Or it can be also written as $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$
So we can substitute the given data.
This gives,
$\Rightarrow$$\cos 60^\circ = \dfrac{{{3^2} + {c^2} - {4^2}}}{{2 \times 3 \times c}}$
Simplifying we get,
$\Rightarrow$$\cos 60^\circ = \dfrac{{9 + {c^2} - 16}}{{6c}}$
$ \Rightarrow \cos 60^\circ = \dfrac{{{c^2} - 7}}{{6c}}$
We know, $\cos 60^\circ = \dfrac{1}{2}$.
Substituting this we get,
$\Rightarrow$$\dfrac{1}{2} = \dfrac{{{c^2} - 7}}{{6c}}$
Cross-multiplying we get,
$\Rightarrow$$2({c^2} - 7) = 6c$
Dividing both sides by $2$ we have,
$\Rightarrow$${c^2} - 7 = 3c$
$ \Rightarrow {c^2} - 3c - 7 = 0$
Thus we get the equation for which $c$ is the root.

$\therefore $ The answer is option A.

Note: Cosine rule can be used in cases if three sides are given or two sides and the included angle are given. There are three variables in the equation: three sides and one angle. So knowing three of them, we can solve for the fourth.