这是一份umass麻省大学 MATH 103作业代写的成功案例
问题 1.
Proof
Drop a perpendicular $h$ from $C$ to $A B$ at $M$.
Then
and
$$
\begin{aligned}
&h^{2}=b^{2}-\overline{A M}^{2}, \
&h^{2}=a^{2}-\overline{M B}^{2} .
\end{aligned}
$$
证明 .
Equating, we have:
$$
a^{2}-\overline{M B}^{2}=b^{2}-\overline{A M}^{2},
$$
or:
$$
a^{2}=b^{2}+\overline{M B}^{2}-\overline{A M}^{2}
$$
Then, since
and
$$
\begin{aligned}
&M B=c-A M \
&=c^{2}-2 c(A M)+\overline{A M}^{2},
\end{aligned}
$$
MATH103 COURSE NOTES ：
Solve the triangle; given $a=20.63, b=34.21, c=40.17$.
We have:
$$
\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c} .
$$
By the table of squares,
$$
\begin{aligned}
&a^{2}=425.6 \
&b^{2}=1171 \
&c^{2}=1614
\end{aligned}
$$
(The interpolation is exactly the same as in logarithms.)