computational with angle wanted, <ABC=45^o, <BAP = 15^o, PC=2, TX=\sqrt3

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computational with angle wanted, <ABC=45^o, <BAP = 15^o, PC=2, TX=\sqrt3

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#1 h Dec 19, 2020, 7:13 PM

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Let $ABC$ be a triangle with $\angle ABC = 45^o$ . Point $P$ lies on the side BC that $PC=2$ units and $\angle BAP = 15^o$ . If the line segment of the tangent drawn from point $B$ to the circumcircle of the triangle $APC$ is $\sqrt3$ units, find the measure of the $\angle ACP$ .

This post has been edited 1 time. Last edited by parmenides51, Dec 19, 2020, 7:18 PM

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#2 h Jan 19, 2021, 7:25 AM

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By the power of a point, we have $(\sqrt3)^2=PB\cdot BC$ , since $BC=BP+PC$ , we get the following:
$PB\cdot (PB+2)=3\implies PB=1$ Thus, $BC=3$ . Also, we have $\sin{30^{\circ}}=2\sin{15^{\circ}}\cos{15^{\circ}}=2\sin{15^{\circ}}\sqrt{1-\sin^2{15^{\circ}}}=$ Thus, $x(\sqrt{1-x^2})=\frac14$ , where $x=\sin{15^{\circ}}$ .
$x^2(1-x^2)=\frac{1}{16}$ Solving we have $x^2=\frac{1\pm\sqrt{1-\frac14}}{2}=\frac{2\pm\sqrt{3}}{4}$ . Notice that positive value is false, since $\sin$ is increasing on $(0,90^{\circ})$ . Thus, $x^2=\frac{2-\sqrt{3}}{4}$ , or $x=\pm\frac{\sqrt{2-\sqrt{3}}}{2}$ , negative is obviously false, thus $x=\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}$ .
By the law of sines, we have $\frac{AP}{\sin{45^{\circ}}}=\frac{PB}{\sin{15^{\circ}}}\implies AP=\frac{\sqrt{2}}{2}\cdot \frac{2\sqrt{2}}{\sqrt{3}-1}\cdot 1=\frac{2}{\sqrt{3}-1}=\sqrt{3}+1$ Therefore, $AC^2=CP^2+AP^2-2CP\cdot AP\cdot \cos{60^{\circ}}=4+4+2\sqrt{3}-2(\sqrt{3}+1)=6\implies AC=\sqrt{6}$
Now we see that $ACP\sim BCA$ , because
$\frac{AC}{CP}=\frac{BC}{AC}\Longleftrightarrow AC^2=CP\cdot BC \Longleftrightarrow (\sqrt{6})^2=2\cdot 3$ Therefore, $\angle CAB=\angle APC=60^{\circ}$
Thus, $\angle CAP=\angle CAB-\angle PAB=60^{\circ}-15^{\circ}=45^{\circ}$ ,
therefore $\angle BCA=180^{\circ}-\angle APC-\angle CAP=180^{\circ}-60^{\circ}-45^{\circ}=\boxed{75^{\circ}}.$

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#3 h Jan 19, 2021, 5:49 PM

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parmenides51 wrote:

Let $ABC$ be a triangle with $\angle ABC = 45^o$ . Point $P$ lies on the side BC that $PC=2$ units and $\angle BAP = 15^o$ . If the line segment of the tangent drawn from point $B$ to the circumcircle of the triangle $APC$ is $\sqrt3$ units, find the measure of the $\angle ACP$ .

Let $T$ be the point of tangency then By Tangent-Secant Theorem we can easily get $PB=1$ as $BT=\sqrt{3}$ Now take a point $H\in BP$ Such that $BP\perp CH$ Now as $PC=2$ Then $PH=1=PB\implies \angle HBC=30^\circ=\angle HCB=\angle BHP$ and also as $\angle BAP=15^\circ$ so $BH=HA$ Hence $H$ is circumcenter of $\triangle ABC$
Hence $\boxed{\angle ACP=75^\circ}$ $\blacksquare$

This post has been edited 1 time. Last edited by Maths_1729, Jan 19, 2021, 5:51 PM

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