$$
\begin{array}{l}
\alpha+\beta=10, \alpha \beta=20 \\
a=\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{10}{20}=\frac{1}{2} \\
\sin \theta=\frac{1}{2} \\
\Rightarrow \theta=30^{\circ} \\
b=10 \cos 60^{\circ}=5
\end{array}
$$

Reference: 1984 FI2.3
Put $x=b=5, y=c$ into the line: $2 c=5+15$
$$
c=10
$$

Reference: 1985 FG10.2, 1986 FG7.3, 1987 FSI.1, 1988 FG9.3
$$
\begin{array}{l}
x^{2}-10 x+40 \equiv(x-5)^{2}+15 \\
k=-5, d=15
\end{array}
$$
I2.1 If $\alpha, \beta$ are roots of $x^{2}-10 x+20=0$, find $a$, where $a=\frac{1}{\alpha}+\frac{1}{\beta}$.
I2.2 If $\sin \theta=a\left(0^{\circ}<\theta<90^{\circ}\right)$, and $10 \cos 2 \theta=b$, find $b$.
I2.3 The point $A(b, c)$ lies on the line $2 y=x+15$. Find $c$.
I2.4 If $x^{2}-c x+40 \equiv(x+k)^{2}+d$, find $d$.