Unable to find all angles of a trigonometric triangle

Question

I have been stuck on this problem for a while now. And have gotten close to slowing it but I am unsure what small mistake I am doing.

This is the question [Here is the Question, it is in Norwegian but translated it says "In the $\triangle ABC$ $AC=8$, $\angle B=30^\circ$ and the length of the Normal $BP$ from $B$ thru the line $AC$ is $6$. Find the angles of $ABC$"[1]

https://i.stack.imgur.com/gFieI.png

[Solution] [2]: https://i.stack.imgur.com/GB3jl.png

This has been my idea so far:

$sin(30) = \frac 8x, x= 16$ such that I find the length of line CB

$$\sin(c) = \dfrac{6}{16}, x= 0.365$$

$$\sin^{-1}(\sin(0.366)) = 22.02$$

I don't bother to do anything more since finding the other angel will of course result in the wrong result

Thank you for the help in advance

Answer 1

Let $AP = x$.
Then $AB = \sqrt{ x^2 + 36}$ and $CB = \sqrt{ x^2 + 16x + 100 }$.

Consider angles $\angle CBP, \angle ABP$.
$\sin \angle CBP = \frac{8+x} { \sqrt{ x^2 + 16x + 100 }}, \cos \angle CBP = \frac{6}{ \sqrt{ x^2 + 16x + 100 } }$.
$\sin \angle ABP = \frac{x}{\sqrt{x^2 + 36 } }, \cos \angle ABP = \frac{6}{\sqrt{x^2 + 36 } }$.

Hence, $ \frac{1}{2} = \sin 30^\circ = \sin (\angle CBP - \angle ABP) = \frac{8+x} { \sqrt{ x^2 + 16x + 100 }} \times \frac{6}{\sqrt{x^2 + 36 } } - \frac{6}{ \sqrt{ x^2 + 16x + 100 } } \times \frac{x}{\sqrt{x^2 + 36 } }$

This gives us $(x^2 + 16x + 100)(x^2 + 36) = 96^2$.
Solving the quartic polynomial for $x$, we get $ x = 2(\sqrt{ 12 \sqrt{3} - 5 } - 2 ) \approx 3.95$

Finally, $\angle BCP = \sin^{-1} \frac{ 6}{\sqrt{ x^2 + 16 x + 100 }} \approx 26.7$ as stated.

Answer 2

If it's part of your knowledge base, considering the triangle $ABC$:

$\qquad\qquad$

to determine the variable $x > 0$ it's sufficient to impose:

$$ \overline{A-B} \cdot \overline{C-B} = ||\overline{A-B}||\;||\overline{C-B}||\cos(30°) $$

from which:

$$ x(x+8) + 36 = \sqrt{x^2+36}\;\sqrt{(x+8)^2+36}\;\frac{\sqrt{3}}{2} $$

then squaring both sides and ordering the monomials, we get:

$$ x^{\color{red}{4}} + \color{blue}{16}\,x^3 + 136\,x^2 + 576\,x - 5616 = 0. $$

Now, the general idea is to reduce to a depressed quartic equation of the cubic term:

$$ x = y - \frac{\color{blue}{16}}{\color{red}{4}} \quad \quad \Rightarrow \quad \quad y^4 + 40\,y^2 - 6512 = 0 $$

where luck wants us to obtain a depressed quartic equation also of the linear term!

It's precisely for this fact that it turns out to be an easily solvable quartic equation:

$$ y^2 = -20+48\sqrt{3} \quad \quad \Rightarrow \quad \quad x = 2\sqrt{12\sqrt{3}-5}-4 $$

from which the required angle:

$$ \widehat{ACB} = \arctan\left(\frac{6}{x+8}\right) \approx \boxed{26.7°}\,. $$

---

Otherwise, set $\widehat{ACB} \equiv \gamma$, applying the law of sines to triangles $ABC$ and $ABP$:

$$ \frac{\sin\gamma}{\overline{AB}} = \frac{\sin(30°)}{8}, \quad \quad \quad \frac{\sin(90°)}{\overline{AB}} = \frac{\sin(\gamma+30°)}{6} $$

from which:

$$ \frac{6}{\overline{AB}} = \frac{\sqrt{3}}{2}\,\sin\gamma + \frac{1}{2}\,\cos\gamma = \frac{\sqrt{3}}{2}\,\frac{\overline{AB}}{16} + \frac{1}{2}\sqrt{1-\left(\frac{\overline{AB}}{16}\right)^2} $$

i.e

$$ \left(\frac{6}{\overline{AB}}-\frac{\sqrt{3}}{2}\,\frac{\overline{AB}}{16}\right)^2 = \left(\frac{1}{2}\sqrt{1-\left(\frac{\overline{AB}}{16}\right)^2}\right)^2 $$

where by expanding the squares and rearranging the monomials, we get:

$$ \overline{AB}^4 - 32\left(3\sqrt{3}+2\right)\overline{AB}^2 + 9216 = 0 \quad \Rightarrow \quad \overline{AB} = 4\sqrt{3\sqrt{3}+2-\sqrt{12\sqrt{3}-5}} $$

from which the required angle:

$$ \gamma = \arcsin\left(\frac{\overline{AB}}{16}\right) \approx \boxed{26.7°}\,. $$

While the calculations remain horrific, this is a way more befitting of a high school.