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The difference between two angles of a triangle is $24^circ$. The average of the same two angles is $54^circ$ .Which one of the following is the value of the greatest angle of the triangle?









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Option(D) is correct

Let $a$ and $b$ be the two angles in the question, with $a > b$. We are given that the difference between the angles is $24^\circ$.

$\Rightarrow  a – b = 24$.

Since the average of the two angles is $54^\circ$, we have \(\dfrac{(a+b)}{2}=54\)

Solving for $b$ in the first equation yields $b=a–24$, and substituting this into the second equation yields










Now, let $c$ be the third angle of the triangle. Since the sum of the angles in the triangle is $180^\circ$, 


Plugging the previous results into the equation yields $66+42+c=180$. 

Solving for $c$ yields $c=72$

Hence, the greatest of the three angles $a, b$ and $c$ is $c$, which equals $72^\circ$.

Edit: For a quick alternative solution, check comment by Sravan Reddy.

(5) Comment(s)


for these probelm only these shortcut is ok ,but if answer >72 we must follow process


Sorry for typo. It's $180-108=72$ as Divya Pointed it out :)


180-54 = 126 not 72 small correction its 180-108=72 (sum of the angles in the triangle-sum of two angles= third side)

Sunil Singh

multiply 54 by 2 and 180-108 = 72 ans.

Sravan Reddy

Killing it in 5 sec:

Average of two angles is 54 => Sum of two angles is 108 and hence 3rd angle is $180-54 = 72$.

In option's that's the largest value given so no need to check whether any other angle is bigger than this :)