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Question: In a triangle \(ABC\), \(a = 4,b = 3,\angle A = 60^\circ \). Then \(c\) is the root of which of the ...

In a triangle ABCABC, a=4,b=3,∠A=60∘a = 4,b = 3,\angle A = 60^\circ . Then cc is the root of which of the following equations?
A) c2−3c−7=0{c^2} - 3c - 7 = 0
B) c2+3c+7=0{c^2} + 3c + 7 = 0
C) c2−3c+7=0{c^2} - 3c + 7 = 0
D) c2+3c−7=0{c^2} + 3c - 7 = 0

Explanation

Solution

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,
a2=b2+c2−2bccos⁡A{a^2} = {b^2} + {c^2} - 2bc\cos A
Or it can be also written as cos⁡A=b2+c2−a22bc\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}

Complete step-by-step solution:
Given that in the triangle ABCABC, a=4,b=3,∠A=60∘a = 4,b = 3,\angle A = 60^\circ .
We need to find the equation with cc as its root.
Here by conventional notation, a,b,ca,b,c represents the sides opposite to vertices A,B,CA,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,
a2=b2+c2−2bccos⁡A{a^2} = {b^2} + {c^2} - 2bc\cos A
Or it can be also written as cos⁡A=b2+c2−a22bc\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}
So we can substitute the given data.
This gives,
⇒\Rightarrow cos⁡60∘=32+c2−422×3×c\cos 60^\circ = \dfrac{{{3^2} + {c^2} - {4^2}}}{{2 \times 3 \times c}}
Simplifying we get,
⇒\Rightarrow cos⁡60∘=9+c2−166c\cos 60^\circ = \dfrac{{9 + {c^2} - 16}}{{6c}}
⇒cos⁡60∘=c2−76c\Rightarrow \cos 60^\circ = \dfrac{{{c^2} - 7}}{{6c}}
We know, cos⁡60∘=12\cos 60^\circ = \dfrac{1}{2}.
Substituting this we get,
⇒\Rightarrow 12=c2−76c\dfrac{1}{2} = \dfrac{{{c^2} - 7}}{{6c}}
Cross-multiplying we get,
⇒\Rightarrow 2(c2−7)=6c2({c^2} - 7) = 6c
Dividing both sides by 22 we have,
⇒\Rightarrow c2−7=3c{c^2} - 7 = 3c
⇒c2−3c−7=0\Rightarrow {c^2} - 3c - 7 = 0
Thus we get the equation for which cc 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.