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In $\triangle ABC$, we have $\angle BAC = 60^\circ$ and $\angle ABC = 45^\circ$. The bisector of $\angle A$ intersects $\overline{BC}$ at point $T$, and $AT = 24$. What is the area of $\triangle ABC$?

I first labeled every angles, $\measuredangle A=60^{\circ}$, $\measuredangle B=45^{\circ},$ $\measuredangle C=75^{\circ}$. I also labeled $\measuredangle ATB=105^{\circ}$, and $\measuredangle ATC=75^{\circ}$.

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closed as off-topic by user10354138, postmortes, Shailesh, Leucippus, Cesareo Jun 23 at 9:41

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  • $\begingroup$ What have you tried? $\endgroup$ – user10354138 Jun 22 at 19:14
  • $\begingroup$ I tried to draw a diagram, but have no idea where to start. $\endgroup$ – I suck at geometry Jun 22 at 19:16
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    $\begingroup$ @I suck at geometry I solved you problem. If you want to see my solution, show please more your attempts. $\endgroup$ – Michael Rozenberg Jun 22 at 19:20
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    $\begingroup$ @user10354138 I first labeled every angles, <a=60, <b=45, <c=75. I also labeled <ATB=105, and <ATC=75. $\endgroup$ – I suck at geometry Jun 22 at 19:36
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    $\begingroup$ It seems like triangle ACT is an isosceles triangle $\endgroup$ – I suck at geometry Jun 22 at 19:37
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Since $$\measuredangle CTA=45^{\circ}+30^{\circ}=75^{\circ}=\measuredangle C,$$ we obtain: $$AC=AT=24.$$ Now, let $TK$ and $TN$ be altitudes of $\Delta ABT$ and $\Delta ACT$ respectively.

Thus, $$TK=TN=\frac{1}{2}AT=12$$ and $$S_{\Delta ATC}=\frac{12\cdot 24}{2}=144.$$ Also, $$S_{\Delta ABT}=\frac{12\cdot(12+\sqrt{24^2-12^2})}{2}.$$ Can you end it now?

I got the area is equal to $72(3+\sqrt3).$

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  • $\begingroup$ thank you very much, but can you use a way that does not need sin because I haven't learn that yet. $\endgroup$ – I suck at geometry Jun 22 at 19:44
  • $\begingroup$ @I suck at geometry I'll post a solution without law of sines. $\endgroup$ – Michael Rozenberg Jun 22 at 19:46
  • $\begingroup$ I think this problem need to use the 90, 60, 30 relationship, and 45, 45, 90 relationship to solve this. $\endgroup$ – I suck at geometry Jun 22 at 19:50
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    $\begingroup$ @MichaelRozenberg You posted a solution using trigonometry and had to change it because the OP doesn't want to use trigonometry. I suggest to add a "geometry-without-trigonometry" tag so users know how should they answer. If you agree please support this suggestion here. $\endgroup$ – Paracosmiste Jun 23 at 18:21
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As you have discovered, $\triangle ACT$ is isosceles.

$$AC = AT = 24$$

Mark $D$ on $AB$ where $AB\perp CD$. Consider two triangles $\triangle ACD$ and $\triangle BCD$.

$\triangle ACD$ is a $60^\circ-30^\circ-90^\circ$ triangle.

$$AD = 12, CD = 12\sqrt 3$$

$\triangle BCD$ is a $45^\circ-45^\circ-90^\circ$ triangle.

$$BD = CD = 12\sqrt 3$$

The area is then $\frac 12 AB\cdot CD.$

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Clearly $$ \angle ATB=105^\circ, \angle ATB=75^\circ $$ and hence $AC=AT=24$. In $\Delta ABT$, by the Law of Sine, one has $$ \frac{AB}{\sin\angle ATB}=\frac{AT}{\sin\angle ABT} $$ which has $$ AB=\frac{24\sin105^\circ}{\sin45^\circ}. $$ So $$ S_{\Delta ABC}=\frac12 AB\cdot AC\sin\angle BAC=\frac12\frac{24^2\sin105^\circ\sin60^\circ}{\sin45^\circ}=72(3+\sqrt3). $$ Here $$ \sin 105^\circ=\frac{\sqrt6+\sqrt2}{4} $$ is used.

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  • $\begingroup$ Your solution is correct but the OP doesn't want to use trigonometry. I suggest to add a "geometry-without-trigonometry" tag so users know how should they answer. If you agree please support this suggestion here. $\endgroup$ – Paracosmiste Jun 23 at 18:19

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