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Q.

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?

 A.

$45^\circ$

 B.

$60^\circ$

 C.

$66^\circ$

 D.

$72^\circ$

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Solution:
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

\(\dfrac{a+(a+24)}{2}=54\)

\(\dfrac{2a-24}{2}=54\)

$2a−24=54×2$

$2a−24=108$

$2a=108+24$

$2a=132$

$a=66$

Also, 

$b=a−24=66−24=42$.

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

$a+b+c=180$. 

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)


Venki
 ()

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



Sravan
 ()

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



Divya
 ()

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 :)