The question provides a problem where the angles of a triangle are in an arithmetic progression (A.P), and the greatest angle is double the least angle. We are to find the sine of the third angle. Here is the step-by-step solution:
- Let the angles of the triangle be \(a-d\), \(a\), and \(a+d\), where \(a-d\), \(a\), \(a+d\) are in A.P. This can be expressed because the common difference in an A.P is \(d\).
- Given that the greatest angle is double the least angle:
- Simplifying the equation:
- \(a + d = 2a - 2d\)
- Rearrange to find \(d\):
- \(3d = a\)
- \(d = \frac{a}{3}\)
- Now, substitute \(d = \frac{a}{3}\) back into the angle expressions:
- First angle: \(a - d = a - \frac{a}{3} = \frac{2a}{3}\)
- Second angle: \(a\)
- Third angle: \(a + d = a + \frac{a}{3} = \frac{4a}{3}\)
- Using the property of a triangle, the sum of angles is \(180^{\circ}\):
- \(\frac{2a}{3} + a + \frac{4a}{3} = 180\)
- \(\frac{2a + 3a + 4a}{3} = 180\)
- \(\frac{9a}{3} = 180\)
- \(3a = 180\)
- \(a = 60\)
- Therefore, the angles are:
- First angle: \(\frac{2 \times 60}{3} = 40\)
- Second angle: \(60\)
- Third angle: \(\frac{4 \times 60}{3} = 80\)
- The third angle, \(80^\circ\), is the angle of interest. We need to find \(\sin(80^\circ)\):
The closest standard value of sine to \(80^\circ\) is \(\sin(60^\circ)\), which is \(\frac{\sqrt{3}}{2}\). Hence, the sine of the third angle is approximately: \(\frac{\sqrt{3}}{2}\).