a: Xét ΔOHB vuông tại H có \(sinOBH=\dfrac{OH}{OB}=\dfrac{1}{2}\)
nên \(\widehat{OBH}=30^0\)
b: Ta có: ΔOAB cân tại O
=>\(\widehat{AOB}=180^0-2\cdot\widehat{OBH}=120^0\)
Xét ΔOAB có \(\dfrac{AB}{sinAOB}=\dfrac{OA}{sinABO}\)
=>\(\dfrac{R}{sin30}=\dfrac{9}{sin120}=6\sqrt{3}\)
=>\(R=6\sqrt{3}\cdot sin30=6\sqrt{3}\cdot\dfrac{1}{2}=3\sqrt{3}\left(cm\right)\)
- \( OH = \frac{R}{2} \)
- \( AH = \frac{AB}{2} = \frac{9}{2} = 4.5 \) cm.
\[
OA^2 = OH^2 + AH^2
\]
\[
R^2 = \left(\frac{R}{2}\right)^2 + 4.5^2
\]
\[
R^2 = \frac{R^2}{4} + 20.25
\]
\[
R^2 - \frac{R^2}{4} = 20.25
\]
\[
\frac{3R^2}{4} = 20.25
\]
\[
3R^2 = 81
\]
\[
R^2 = 27
\]
\[
R = \sqrt{27} = 3\sqrt{3} \text{ cm}
\]
\[
\sin \angle OHA = \frac{AH}{OA} = \frac{4.5}{R}
\]
\[
\sin \angle OHA = \frac{4.5}{3\sqrt{3}} = \frac{1.5}{\sqrt{3}} = \frac{1.5 \sqrt{3}}{3} = \frac{\sqrt{3}}{2}
\]
\[
\sin 60^\circ = \frac{\sqrt{3}}{2}
\]
Do đó, \( \angle OHA = 60^\circ \).
Vì \( \angle OBH = 90^\circ - \angle OHA \), nên:
\[
\angle OBH = 90^\circ - 60^\circ = 30^\circ
\]
