Question

The smallest angle of the ABC when $a=7,b=4\sqrt{3} and c=\sqrt{13}$, is [MP PET 2003]
A. ${{30}^{o}}$
B. ${{15}^{o}}$
C. ${{45}^{o}}$
D. None of these

Answer 1

Hint: In this question, we have given the angles of a triangle with its side measures. To find the smallest angle use the laws of the cosine of angle relation. Use rationalization to simplify the expression. Lastly, compare the obtained result with the option provided.


Formula used:Laws of cosine for triangle $ABC$ whose length is $a, b$, and $c$ respectively is given by;
$b^2 = a^2 +c^2 − 2ac.\cos B$.


Complete step by step solution:Given sides of $\Delta ABC$ are $a=7,b=4\sqrt{3}$ and $c=\sqrt{13}$.
We know that the smallest angle in a triangle has the shortest opposing side and here the shortest side is $c=\sqrt{13}$.
Therefore,
$\cos C=\dfrac{{{b}^{2}}+{{a}^{2}}-{{c}^{2}}}{2ab}$
Substituting the values we get;
$=\dfrac{(7)^2+(4\sqrt{3})^2-(\sqrt{13})^2}{2\times 7\times 4\sqrt{3}}
\\=\dfrac{49+48-13}{2\times 7\times 4\sqrt{3}}\\
=\dfrac{84}{56\sqrt{3}}\\
=\dfrac{3}{2\sqrt{3}}$
Now rationalize;
$=\dfrac{3}{2\sqrt{3}}\times \dfrac{2\sqrt{3}}{2\sqrt{3}}\\
=\dfrac{6\sqrt{3}}{12}\\
=\dfrac{\sqrt{3}}{2}$
Thus, $\cos C=\dfrac{\sqrt{3}}{2}$
$\Rightarrow C=\cos^{-1} \dfrac{\sqrt{3}}{2}$
$\Rightarrow C={{30}^{o}}$
Thus, the smallest angle is
$\angle C={{30}^{o}}$


Thus, Option (A) is correct.

Note: Keep in mind that here we are finding the smallest angle in a triangle and we know that the smallest angle in a triangle has the shortest opposing side. Here it is C. Using the law of cosine for angle C, students usually make mistakes in taking the wrong angles. Different angles use cosine rules in a different way.