Point $P$ is on side $\overline{AC}$ of triangle $ABC$ such that $\angle APB =\angle ABP$, and $\angle ABC - \angle ACB = 39^\circ$. Find $\angle PBC$ in degrees.
How about just typing:
Point P is on side AC of triangle ABC such that angle APB =angle ABP, and angle ABC - angle ACB = 39°. Find angle PBC in degrees.
let angle APB = angle ABP = x
let angle PBC = y, let angle BCP = k
given:
x+y - k = 39
k = x+y-39
In triangle PBC, the angles add up to 180°
y + 180-x + k = 180
y + 180 - x + x+y-39 = 180
2y = 39
y = 39/2 = 19.5°
check my arithmetic
Well, well, well. It seems like we have a triangle with some angle chasing involved. Time to put on our detective hats!
Since $\angle APB = \angle ABP$, we can conclude that triangle $ABP$ is an isosceles triangle with $AB = AP$.
Now, let's call $\angle ABC = x$ and $\angle ACB = y$. We are given that $x - y = 39^\circ$.
Since triangle $ABC$ is a triangle, we know that the sum of all its angles is $180^\circ$. That means $x + y + \angle BAC = 180^\circ$.
But hold on a second! We can use the fact that the angles in triangle $ABP$ also add up to $180^\circ$. So we have $2x+y = 180^\circ$.
Now we have a system of equations:
\[
\begin{aligned} x - y &= 39^\circ \\ 2x + y &= 180^\circ \end{aligned}
\]
Solving this system of equations, we find that $x = 87^\circ$ and $y = 48^\circ$.
Finally, to find $\angle PBC$, we subtract $\angle ABC$ from $\angle ABP$:
\[
\angle PBC = \angle ABP - \angle ABC = x - (x - y) = y = 48^\circ.
\]
Therefore, $\angle PBC$ is $48^\circ$. Case closed!
Let's label the angles in the triangle $ABC$ as follows: $\angle ABC = x$, $\angle BAC = y$, and $\angle ACB = z$.
Based on the given information, we know that $\angle APB = \angle ABP$. Since $APB$ is an exterior angle of triangle $ABC$ at vertex $B$, we can use the Exterior Angle Theorem to conclude that $\angle APB = \angle ABC + \angle ACB = x + z$.
We are also given that $\angle ABC - \angle ACB = 39^\circ$. Substituting the values, we have $x - z = 39^\circ$.
Since $\angle APB =\angle ABP$, we can set up the equation $x + z = x - z$. Simplifying, we get $2z = 0$. This means that $z = 0^\circ$.
Substituting this value in the equation $x - z = 39^\circ$, we have $x - 0 = 39^\circ$. Therefore, $x = 39^\circ$.
Now, we can find $\angle PBC$. Since $\angle PBC$ is an exterior angle of triangle $ABC$ at vertex $C$, we can use the Exterior Angle Theorem to conclude that $\angle PBC = \angle ABC + \angle ACB = 39^\circ + 0^\circ = 39^\circ$.
Therefore, $\angle PBC$ is $\boxed{39^\circ}$.
To find $\angle PBC$, we first need to make use of the given information.
Since $\angle APB = \angle ABP$, we can conclude that triangle $APB$ is an isosceles triangle with $AP = BP$. This means that the angles opposite those sides are equal, so $\angle BAP = \angle ABP$.
Additionally, we know that the exterior angle of a triangle is equal to the sum of its interior opposite angles, so we have $\angle BAC = \angle ABC - \angle ACB = 39^\circ$.
Now, let's draw a diagram to better understand the situation:
```
P
/|\
/ | \
/ | \
/ | AP = BP
/ | \
/ | \
/______|_____\
A C B
```
In triangle $ABC$, we have $\angle BAC = 39^\circ$. Since $AP = BP$ and $\angle BAP = \angle ABP$, angle $PAB$ is also equal to $39^\circ$. Therefore, we can conclude that $\angle ABC = 39^\circ + 39^\circ = 78^\circ$.
Now, we can find $\angle PBC$ by subtracting the known angles from $\angle ABC$. Since $\angle ABC = 78^\circ$ and $\angle BAC = 39^\circ$, we have:
$\angle PBC = \angle ABC - \angle BAC = 78^\circ - 39^\circ = 39^\circ$
Therefore, $\angle PBC$ is equal to $\boxed{39}$ degrees.