Question

Question 13
The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.

Answer 1

Given that, the angles of a triangle are in AP.
Let us consider $\Delta$ ABC. A, B and C are the three angles which are in AP.
$\therefore B=\frac{A+C}{2}$
$\Rightarrow 2B=A+C...\left(i\right)$
We know that, sum of all interior angles of a $\Delta ABC.={180}^{\circ }$
$\Rightarrow A+B+C={180}^{\circ }$
$\Rightarrow 2B+B={180}^{\circ }[fromeq.(i\left)\right]$
$\Rightarrow 3B={180}^{\circ }$
$\Rightarrow B={60}^{\circ }$
Let us consider A as the greatest angle and C as the least angle.
So, A = 2C[ by condition] ...(ii)
Substituting the values of B and A in eq.(i), we get;
2$\times$60 = 2C + C
$\Rightarrow 120=3C\Rightarrow C={40}^{\circ }$
Put the value of C in eq.(ii), we get;
$A=2\times {40}^{\circ }\Rightarrow A={80}^{\circ }$
Hence, the required angles of the triangle are ${80}^{\circ },{60}^{\circ }and{40}^{\circ }$.