Q.The angles of a triangle are in the ratio $2:3:4$. Express each angle in radian measure.

The three angles of any triangle add up to $180^\circ$, so the ratio parts must together account for that total before each angle can be converted to radians.

Ratio parts $2+3+4=9$; one part $=180^\circ/9=20^\circ$. Angles: $40^\circ, 60^\circ, 80^\circ$, i.e. $\dfrac{2\pi}{9}, \dfrac{\pi}{3}, \dfrac{4\pi}{9}$ radians.

Step 1 — use the angle-sum property of a triangle. The three angles are $2x, 3x, 4x$ for some common multiplier $x$, and $2x+3x+4x=180^\circ$ gives $9x=180^\circ$, so $x=20^\circ$.

Step 2 — find each angle in degrees. $2x=40^\circ$, $3x=60^\circ$, $4x=80^\circ$.

Step 3 — convert each to radians using $1^\circ=\pi/180$: $40^\circ=\dfrac{40\pi}{180}=\dfrac{2\pi}{9}$; $60^\circ=\dfrac{60\pi}{180}=\dfrac{\pi}{3}$; $80^\circ=\dfrac{80\pi}{180}=\dfrac{4\pi}{9}$.

Check (independent route — the three radian angles must also sum to $\pi$). $\dfrac{2\pi}{9}+\dfrac{\pi}{3}+\dfrac{4\pi}{9} = \dfrac{2\pi}{9}+\dfrac{3\pi}{9}+\dfrac{4\pi}{9} = \dfrac{9\pi}{9}=\pi$, exactly matching $180^\circ=\pi$ radians — confirming all three conversions are correct.

Assuming the ratio parts themselves are the angles in degrees (wrongly taking the angles as $2^\circ, 3^\circ, 4^\circ$) instead of first finding the multiplier $x$ from the triangle's angle-sum property; also converting only one angle to radians and leaving the others in degrees.