$\triangle ABC, \angle{ABC} = 90^{\circ}$.
Let $P$ be the point on $BC$ such that $2 \cdot \angle{BAP} = \angle{CAP} = 14 ^{\circ}$
and $Q$ be the point on $AB$ such that $\angle{BCQ} = 23 ^{\circ}$.
Find $\angle{BPQ}$.
My work :
$\angle{APB} = 83 ^{\circ}$
$\angle{BQC} = 67 ^{\circ}$
$\angle{ACQ} = 46 ^{\circ}$
Let $QC \cap AP= U$, $\angle{QUP} = 120 ^{\circ}$
Let's call $\angle BPQ =x$ then at the triangle $PQB$ we get
$$\tan x=\frac{QB}{PB}$$
but, at the triangle $CQB$ we get:
$$\tan 23°=\frac{QB}{CB}\to QB=CB\cdot \tan 23°$$
and from the triangle $ABP$ we have:
$$\tan 7°=\frac{PB}{AB}\to PB=AB\cdot \tan 7°$$
so,
$$\tan x=\frac{CB}{AB}\cdot \frac{\tan 23°}{\tan 7°}$$
Now from the triangle $ABC$ we have
$$\tan 21°=\frac{CB}{AB}$$
and then
$$\tan x=\frac{\tan 21°\cdot\tan 23°}{\tan 7°}\to x=\arctan \left(\frac{\tan 21°\cdot\tan 23°}{\tan 7°}\right)= 53°$$
Again, no need for trigonometric formulas and clumsy calculations.
Let $D$ be the intersection point of $AP$ and $CQ$. By direct angle chasing, one can calculate that $$\angle \, ACD = \angle \, ACQ = 46^{\circ} = 2 \cdot 23^{\circ} = 2 \cdot \angle \, BCQ = 2 \cdot \angle \, PCD$$ Moreover, observe that $$\angle \, CAD = \angle \, CAP = 14^{\circ} = 2 \cdot 7^{\circ} = 2 \cdot \angle \, BAP = 2 \cdot \angle \, QAD$$ Let $I$ be the intersection point of the interior angle bisectors through vertices $A$ and $C$ of triangle $ACD$. Therefore, $DI$ is the interior angle bisector of angle $\angle \, ADC$. Again, some direct angle chasing leads to the conclusion that $\angle \, ADC = 120^{\circ}$ and $\angle \, QDA = \angle \, PDC = 60^{\circ}$.
Now, observe that
$$\angle \, QAD = 7^{\circ} = \frac{1}{2} \cdot 14^{\circ} = \frac{1}{2} \cdot \angle CAD = \angle \, IAD$$
$$\angle \, PCD = 23^{\circ} = \frac{1}{2} \cdot 46^{\circ} = \frac{1}{2} \cdot \angle \, ACD = \angle \, ICD$$
Furthermore, $$\angle \, IDA = \frac{1}{2} \cdot \angle \, ADC = \frac{1}{2} \cdot 120^{\circ} = 60^{\circ} = \angle \, QDA$$
$$\angle \, IDC = \frac{1}{2} \cdot \angle \, ADC = \frac{1}{2} \cdot 120^{\circ} = 60^{\circ} = \angle \, PDC$$ Consequently the triangles $ADQ$ and $ADI$ are congruent, as well as triangles $CDP$ and $CDI$ are also congruent. Therefore, $DQ = DI$ and $DP = DI$. Hence $DQ = DI = DP$ and so triangle $DPQ$ is isosceles with $\angle \, PDQ = 120^{\circ}$. Thus
$$\angle \, QPA = \angle \, QPD = 30^{\circ}$$ By angle chasing, $\angle \, BPA = 83^{\circ}$ and for that reason
$$\angle \, BPQ = \angle \, BPA - \angle \, QPA = 83^{\circ} - 30^{\circ} = 53^{\circ}$$
Hint: $BP = AB \times tan (\angle{BAP})$ and $AB = AC \times cos (\angle{BAC})$.
