\(\Rightarrow R^2+A^2=B^2 \ldots (1)\)
\(R=10\)
Also \(\tan 30^{\circ}=\frac{\text { Perpendicular }}{\text { Base }}\)
\(\frac{1}{\sqrt{3}}=\frac{R}{A}\)
From equation \((1)\) \(A=10 \sqrt{3}\)
\((10)^2+(10 \sqrt{3})^2=B^2\)
\(B=20\)
[$\sqrt{3}=1.7, \sqrt{2}=1.4$ , $\hat{{i}}$ અને $\hat{{j}}$ એ ${x}, {y}$ અક્ષની દિશાના એકમ સદીશ છે.$]$