# In $\triangle ABC$, we have $\angle BAC = 60^\circ$ and $\angle ABC = 45^\circ$… [closed]

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 $$\measuredangle ATC=75^{\circ}$$.

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

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• What have you tried? – user10354138 Jun 22 at 19:14
• I tried to draw a diagram, but have no idea where to start. – I suck at geometry Jun 22 at 19:16
• @I suck at geometry I solved you problem. If you want to see my solution, show please more your attempts. – Michael Rozenberg Jun 22 at 19:20
• @user10354138 I first labeled every angles, <a=60, <b=45, <c=75. I also labeled <ATB=105, and <ATC=75. – I suck at geometry Jun 22 at 19:36
• It seems like triangle ACT is an isosceles triangle – I suck at geometry Jun 22 at 19:37

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 $$72(3+\sqrt3).$$

• thank you very much, but can you use a way that does not need sin because I haven't learn that yet. – I suck at geometry Jun 22 at 19:44
• @I suck at geometry I'll post a solution without law of sines. – Michael Rozenberg Jun 22 at 19:46
• I think this problem need to use the 90, 60, 30 relationship, and 45, 45, 90 relationship to solve this. – I suck at geometry Jun 22 at 19:50
• @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. – Paracosmiste Jun 23 at 18:21

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 $$\frac 12 AB\cdot CD.$$

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.

• 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. – Paracosmiste Jun 23 at 18:19