Two coordinates, two angles, and the third coordinate

Question

Let $A$, $B$ and $C$ be points on a two-dimensional coordinate system. Assume $A=(0,1), B=(0,5)$, angle $\alpha$ of $A$ is 47 degrees, and angle $\beta$ of $B$ is 80 degrees.

Calculate the coordinates of C to one decimal place.

To show effort: I can use the law of sines to find out that $AC = 4.9$, and $BC = 3.7$. But I have no idea how to pinpoint coordinate $C$ without using precise tools, which, I assume, isn't the point. How do I solve this problem?

Answer 1

• Before going into detail, a general remark: Given points $O:=\begin{pmatrix}0\\0\end{pmatrix}$, $E:=\begin{pmatrix}1\\0\end{pmatrix}$ and an angle $\varphi:=\angle EOP$ ($\varphi$ is the oriented angle from $E$ to $P$ centered in $O$), the point $P$ has the coordinates $\begin{pmatrix}\cos \varphi\\\sin\varphi\end{pmatrix}$. Moreover, the distance $\overline{OP}$ is $1$. These are very central facts in geometry.
• Therefore, the distance between points $O$ and $t\cdot\begin{pmatrix}\cos \varphi\\\sin\varphi\end{pmatrix}$ is $t$.
• Now lets define $A':=O$, $B':=B-A$ and $C':=C-A$. So the triangle $A'B'C'$ is only a translated version of the triangle $ABC$.
  • Now lets calculate the angle $\varphi_A:=\angle EOC'$. This can be rewritten as $\varphi_A=\angle EOB'+\angle B'OC'$. We know that $\angle EOB'=90°$ (take a look at the position of the points $A$ and $B$!) and we know that $\angle B'OC'=47°$. So $\varphi_A=137°$.
  • Therefore, the point $C'$ must fulfill the equation $C'=t\begin{pmatrix}\cos \varphi_A\\\sin\varphi_A\end{pmatrix}$ for a yet unknown parameter $t$ that represents its distance to $O$. By adding $A$ on both sides, we get $C=A+C'=A+t\begin{pmatrix}\cos \varphi_A\\\sin\varphi_A\end{pmatrix}$.
• With the same method as above, you can get the equation $C=B+s\begin{pmatrix}\cos \varphi_B\\\sin\varphi_B\end{pmatrix}$, where $\varphi_B:=270°-80°=190°$.
• One method for solving the original problem is solving the equation
  $A+t\begin{pmatrix}\cos \varphi_A\\\sin\varphi_A\end{pmatrix}=B+s\begin{pmatrix}\cos \varphi_B\\\sin\varphi_B\end{pmatrix}$.
  All values except $s$ and $t$ are known, and by splitting the equation into an equation for the $x$-coordinate and an equation for an $y$-coordinate, you get a system of two linear equations with two unknowns, which is solvable.
• But there is an easier method: We already said that $t=\overline{AC}$. So you can simple calculate $\overline{AC}$ (which you actually already did).
• The solution is $C=\begin{pmatrix}0\\1\end{pmatrix}+t\begin{pmatrix}\cos 137°\\\sin 137°\end{pmatrix}\approx\begin{pmatrix}-3{.}6\\4{.}4\end{pmatrix}$