# Difficult Geometry & Mensuration Solved QuestionAptitude Discussion

 Q. One of the angles of a triangle is $60^\circ$ and length of its two sides is 6 cm and 7 cm. The length of the third side of the triangle is:
 ✖ A. $\sqrt{43}$ ✖ B. $3+\sqrt{22}$ ✔ C. $\sqrt{43}$ or $3+\sqrt{22}$ ✖ D. None

Solution:
Option(C) is correct

Case-I:

When the unknown side faces $60^\circ$angle. From cosine rule

$\cos 60^\circ=\dfrac{6^2+7^2-x^2}{2\times 6\times7}$

$x=\pm \sqrt{43}$

$x$ cannot be equal to $x=- \sqrt{43}$

Case-II:

When the side length 7 faces $60^\circ$ angle, From cosine rule

$\cos 60^\circ=\dfrac{6^2+x^2-7^2}{2\times 6\times x}$

$x^2-6x-13=0$

$x= 3\pm \sqrt{22}$

$x$ cannot be equal to

$3-\sqrt{22}$

So, $x=3+\sqrt{22}$ solving the similar way we realise that $60^\circ$angle cannot be opposite to the side of length 6 cm.

Edit: Thank you Vikas, for pointing out, corrected the typo to change the value of $x$ from $43$ to $\pm \sqrt{43}$.

## (2) Comment(s)

Sajawal
()

i think there is an easy method to solve this problem

Simply we know that the length of one side of the traingle can not be bigger then the lenght of other two sides and cannot to less then the difference of the above mentioned two other sides

1

Vikas
()

in case 1 , the value of x is 43 is incorrect as x must be +√43 and -√43