In the given figure, $AB$ is a tangent to the circle centered at $O$. If $OA = 6$ cm and $\angle OAB = 30°$,…

Question

In the given figure, $AB$ is a tangent to the circle centered at $O$. If $OA = 6$ cm and $\angle OAB = 30°$, then the radius of the circle is :

Accepted Answer

## Model Answer **(b) $3\sqrt{3}$ cm** Since OB ⊥ AB (radius ⊥ tangent), in right △OAB: $\sin 30° = \dfrac{OB}{OA}$, so $OB = 6 	imes \dfrac{1}{2}$... Wait — $\sin(\angle OAB) = \dfrac{OB}{OA}$, i.e., $\sin 30° = \dfrac{r}{6}$, giving $r = 3$ cm. **(a) 3 cm** ## Explanation In right △OAB, ∠OBA = 90° (radius ⊥ tangent). The angle at A is 30°, and OA (hypotenuse) = 6 cm. Using sin: $\sin 30° = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{OB}{OA} = \frac{r}{6}$. Since $\sin 30° = \frac{1}{2}$, we get $r = 3$ cm. The correct answer is **(a) 3 cm**. A common mistake is using tan or cos instead of sin — remember OB is the side **opposite** to ∠OAB, and OA is the hypotenuse.

Mathematics — Standard (041) — CBSE Class 10 board question