Question

In a triangle $\mathrm{\Delta }ABC$ if $a=2,b=3,c=4$ then $\cos (A)=?$
(A) ${\displaystyle \frac{2}{9}}$
(B) ${\displaystyle \frac{5}{7}}$
(C) ${\displaystyle \frac{6}{7}}$
(D) ${\displaystyle \frac{7}{8}}$

Answer 1

Hint: When we construct the triangle, if we assign the names $A,B,C$ to the corners, then the side that lies opposite to each angle should be named with that letter. If a corner is named as $A$, then the side opposite this corner angle is the side ‘$a$’. Looking at the given details, the easiest method would be to use the law of cosines to calculate the required value.

Complete step by step solution:
The given details tell us that;
We have a triangle $\mathrm{\Delta }ABC$.
The side length of side $a=2$
Side length $b=3$
And side length $c=4$
We need to find $\cos (A)=?$
Find the relationship between sides and angles to find the required value.
Now we need to remember that when we are given sides of a triangle but the placement of the sides are not mentioned, it is implied that the side is the side of the opposite angle with the same name. So here we are given values of sides $a,b,c$ this implies that they lie opposite to the angles $A,B,C$ respectively.
The given details have been illustrated in the figure below.

Now to find the cosine value of an angle for a triangle whose three sides are given, we make use of the formula of Law of Cosines.
This law gives us the formula:
$\Rightarrow a^2=b^2+c^2-2bc\times \cos (A)$
(This formula can be modified according to the value of cosine we require.)
Substituting the values we have in the equation of Law of Cosines:
$\Rightarrow 2^2=3^2+4^2-2\times 3\times 4\times \cos (A)$
$\Rightarrow 2\times 3\times 4\times \cos (A)=3^2+4^2-2^2$
Isolating the value we need to find:
$\Rightarrow \cos (A)={\displaystyle \frac{3^2+4^2-2^2}{2\times 3\times 4}}$
Solving we get:
$\Rightarrow \cos (A)={\displaystyle \frac{25-4}{24}}$
$\Rightarrow \cos (A)={\displaystyle \frac{21}{24}}$
Simplifying further:
$\Rightarrow \cos (A)={\displaystyle \frac{7}{8}}$
So we have found the value of $\cos (A)$ to be ${\displaystyle \frac{7}{8}}$.
If we take a look at the options, we see that the options (A), (B) and (C) that have the values ${\displaystyle \frac{2}{9}}$, ${\displaystyle \frac{5}{7}}$ and ${\displaystyle \frac{6}{7}}$ respectively, are evidently incorrect options. This is because after solving the given problem using the given values of sides and Law of Cosines the answer we got was ${\displaystyle \frac{7}{8}}$ and none of the options among (A), (B) and (C) have this value.
Therefore the right answer is option (D) ${\displaystyle \frac{7}{8}}$.

Note:
Above we are using the law of cosines because we have the values of all the three sides of a triangle. In an alternate case if we are given that one of the angles are right angled or ${90}^{\circ }$, we can use the law of hypotenuse to find the trigonometric values . This is because any side that lies directly opposite to the right angle is called a hypotenuse and by law of hypotenuse the squares of sums of the other two sides give the squared value of hypotenuse.