Question

The angles of a triangle are in A.P and the greatest angle is double the least angle, then sine of the third angle is

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{2}$
• D. 0

Hint:

A crucial geometric shortcut: If the three angles of a triangle are in Arithmetic Progression, the middle angle is always exactly $60^\circ$, regardless of the other conditions. The question asks for the "third angle" (implicitly the middle one), so you can often jump straight to $\sin(60^\circ) = \sqrt{3}/2$ without even calculating $d$!

Answer 1

The Correct Option is A

Step 1: Let the Angles in A.P.
Let the three angles be: \[ (a - d), \; a, \; (a + d) \]
Step 2: Sum of Angles of Triangle
\[ (a - d) + a + (a + d) = 180^\circ \] \[ 3a = 180^\circ \Rightarrow a = 60^\circ \]
Step 3: Given Condition
Greatest angle = double the least angle: \[ a + d = 2(a - d) \] Substitute $a = 60^\circ$: \[ 60 + d = 2(60 - d) \] \[ 60 + d = 120 - 2d \] \[ 3d = 60 \Rightarrow d = 20^\circ \]
Step 4: Find All Angles
\[ \text{Least angle} = 60 - 20 = 40^\circ \] \[ \text{Middle angle} = 60^\circ \] \[ \text{Greatest angle} = 60 + 20 = 80^\circ \]
Step 5: Required Value
The third (middle) angle is $60^\circ$: \[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]
Step 6: Final Answer
\[ \boxed{\frac{\sqrt{3}}{2}} \]