Question

Angles of a triangle are in A.P. and the greatest angle is double the least. Find the angles in circular measures.
A) 40, 60, 80
B) 40, 60, 50
C) 40, 70, 80
D) 20, 60, 80

Answer 1

Hint:
Here we have to calculate the angles of a triangle. For that, we will first assume the three angles which will form an A.P with the same common difference. Then we will find the sum of all three angles and we will equate the sum with ${180}^{\circ }$ as the sum of all the angles of a triangle is 180o. We will equate the greatest angle with twice the smallest angle. By solving these two equations, we will get the value of all the angles.

Complete step by step solution:
Let all the angles of a triangle which are in A.P be $a-d,a,a+d$
We know the sum of all the angles of a triangle is 180o.
Now, we will find the sum of all the angles that we have assumed.
Sum of angles $=a-d+a+a+d=3a$
We will equate the sum with 180o.
$\Rightarrow 3a={180}^{\circ }$
Now, we will divide 180o by 3.
$\Rightarrow a={\displaystyle \frac{{180}^{\circ }}{3}}=60$
Putting value in all the angles. The angles become $60-d,60,60+d$
We will now equate the greatest angle with twice the smallest angle.
Here the greatest angle is twice the smallest angle.
Therefore,
$\Rightarrow 60+d=2(60-d)$
Now, we will use the distributive law of multiplication here.
$\Rightarrow 60+d=120-2d$
Taking similar terms to the same sides, we get.
$\Rightarrow 3d=60$
Now, we will divide 60 by 3.
$\Rightarrow d={\displaystyle \frac{60}{3}}=20$
Putting value of d in all the angles, we get
$\begin{array}{rl}& \Rightarrow 60-d=60-20={40}^{\circ }\\ & \Rightarrow 60+d=60+20={80}^{\circ }\end{array}$

Therefore, the angles are
${40}^{\circ },{60}^{\circ }\mathrm{\&}{80}^{\circ }$
Thus, the correct option is A.

Note:
We have used distributive law of multiplication here for finding the product of 2 and $60-d$
Let there are four terms $a,b\mathrm{\&}c$ then according to distributive law of multiplication $a(b+c)=a.b+a.c$
Here we have assumed angles as$a-d,a,a+d$. We can also take the angles $a,a+d\mathrm{\&}a+2d$ instead of $a-d,a,a+d$. As both of them are forming an A.P but we generally prefer $a-d,a,a+d$as it makes the solution easy and short.