# Angles on a point inside a triangle

Let $ABC$ be an isosceles triangle with $AB=AC$ and $∠BAC = 100$. A point $P$ inside the triangle $ABC$ satisfies that $∠CBP=35$ and $∠PCB= 30$. Find the measure, in degrees, of angle $∠BAP$. Attached is the figure of the triangle

I tried to Angle Chase but it seemed true for all values of $BAP$. I then tried using the sine law. In the triangle $PBC$, We have $$\frac{PB \sin(35)}{\sin(30)}= PC$$

Trying it with triangles $APB$ ($x= BAP$) and the fact that they are isosceles

$$\frac{PC\sin(x+70)}{\sin(100-x)}=\frac{PB\sin(175-x)}{\sin (x)}$$ Which becomes,

$$\frac{\sin(35)\sin(x+70)}{\sin(30)\sin(100-x)} = \frac{\sin(175-x)}{\sin(x)}$$

Where in I don't know how to solve it. Other methods are welcome.

• Is the sidelength of $AB=AC$? Jul 17, 2018 at 10:45
• Yes, edited in m Jul 17, 2018 at 10:47

Let $M$ be the midpoint of $BC.$ Let $Q$ be the intersection of $CP$ with $AM.$

Note that $\angle ABC = \frac12(180^\circ - 100^\circ) = 40^\circ$ and therefore $$\angle PBA = \angle ABC - \angle PBC = 40^\circ - 35^\circ = 5^\circ.$$

Since $Q$ is on the perpendicular bisector of $BC,$ triangle $\triangle BQC$ is isoceles with $BQ = CQ.$ Moreover, \begin{align} \angle QBC &= \angle QCB = 30^\circ,\\ \angle PBQ &= \angle PBC - \angle QBC = 35^\circ - 30^\circ = 5^\circ,\\ \angle PQB &= \angle QCB + \angle QBC = 60^\circ,\\ \angle PQA &= \angle MQC = 90^\circ - \angle QCB = 60^\circ.\\ \end{align}

In summary, $\angle PBA = \angle PBQ$ and $\angle PQA = \angle PQB.$ That is, the rays $BP$ and $QP$ are the angle bisectors of angles $\angle ABQ$ and $\angle AQB$ of triangle $\triangle ABQ.$ The angle bisectors of a triangle are concurrent, hence $AP$ is an angle bisector of $\angle BAQ.$ But $\angle BAQ = 50^\circ,$ so $\angle BAP = \frac12\angle BAQ = 25^\circ.$

# Solution

Let $$M$$ be the midpoint of $$BC$$. $$AM$$ intersects $$CP$$ at $$Q$$. By the symmetry properties, we may have $$\angle QBC=\angle QCB=\angle PCB=30^o.$$ Thus, $$\angle PBQ=\angle PBC-\angle QBC=35^o-30^o=5^o.$$ But $$\angle ABP=\angle ABC-\angle PBC=\frac{1}{2}(180^o-\angle BAC)-\angle PBC=5^o.$$

Hence, $$\angle PBQ=\angle ABP,$$ namely, $$BP$$ bisects $$\angle ABQ.$$ Morover, since $$\angle AQP=\angle MQC=\angle MQB=90^o-\angle QCM=90^o-\angle PCB=60^o,$$ and $$\angle PQB=2\angle BCQ=2\angle BCP=60^o,$$hence $$\angle AQP=\angle PQB,$$namely, $$QP$$ bisects $$\angle AQB.$$ To sum up, $$P$$ is the incenter of $$\triangle ABQ$$. Thus, $$AP$$ bisects $$\angle BAQ$$. It follows that $$\angle BAP=\frac{1}{2}\angle BAM= \frac{1}{4}\angle BAC=25^o.$$

Let $x=\angle APB$, $a=AB=AB$, $b=AP$. We have $$\frac{\sin x}{a}=\frac{\sin 5^\circ }{b}$$ and $$\frac{\sin \left(245^{\circ} - x\right)}{a}=\frac{\sin 10^\circ }{b}$$

We have then: $$\frac{\sin \left(245^{\circ} - x\right)}{\sin x}=\frac{\sin 10^\circ }{\sin 5^\circ }$$ Using the formula $$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha$$ we obtain: $$\cot x = \frac{\sin 10^\circ }{\sin 245^\circ \sin 5^\circ } + \cot 245^\circ$$ Thus $$x = \text{arccot}\left(\frac{\sin 10^\circ }{\sin 245^\circ \sin 5^\circ } + \cot 245^\circ\right)$$ After wolframAlpha we have $$x=150^{\circ}$$ Now it's easy to show, that $$\angle PAB = 25^\circ$$

• Sorry, but the answer is 25 degrees, source = chiuchang.org/wp-content/uploads/sites/2/2018/01/… Jul 17, 2018 at 11:27
• @Mason You put the arguments of trigonometric functions in radians, not degrees. Jul 17, 2018 at 11:36
• @Mason wolframalpha.com/input/?i=Arccot(Sin(Pi*10%2F180)%2FSin(Pi*5%2F180)%2BCot(Pi*225%2F180))*180%2Fpi Jul 17, 2018 at 11:39
• Your link has 225 but your post has a 245. I think that this is still evaluating to be close to 22 degrees where as OP claims the solution to be 25. Jul 17, 2018 at 11:45
• I've missed $\sin 245^\circ$. Now ti works fine. Jul 17, 2018 at 12:30