In triangle ABC, we have <BAC = 60 degrees and <ABC = 45 degrees. The bisector of <A intersects line BC at point T, and AT = 24. What is the area of triangle ABC?
How should i approach this?
i got AC=24 and AB= 32.78. This gave me 340.7 which was wrong. What should i do?
I got the same results, and I stand by them
Lol you literally copied the question from AoPS... this is against the honor code but whatever... u can ask stuff on the message board.
and you also copied the answer bro... if ya gonna plagiarize at least do it well
##### is correct.
Since the angles in triangle $ABC$ must add up to $180^\circ$, we have $\angle ACT = 75^\circ$.
Similarly, $\angle ATC = 75^\circ$, because the angles in $\triangle ACT$ also add up to $180^\circ$. Two of the angles in $\triangle ACT$ are equal, so it is an isosceles triangle and $AC = AT = 24$.
Seeing the $30^\circ$, $45^\circ$, and $60^\circ$ angles, we draw $\overline{CH}$ of triangle $ABC$ to form a 30-60-90 triangle and a 45-45-90 triangle:
Since $ACH$ is a 30-60-90 right triangle with hypotenuse $AC = 24$, we have $CH = 12\sqrt{3}$ and $AH = AC/2 =12$. $\overline{BH}$ and $\overline{CH}$ are the legs
of 45-45-90 triangle $BCH$ so $BH = CH =12\sqrt{3}$. Therefore, $AB = AH + HB = 12 +12\sqrt{3}$ and
\[[ABC]= \frac{1}{2}(AB)(CH)= \frac{1}{2}(12 + 12\sqrt{3})(12\sqrt{3}) =\boxed{ 216 + 72\sqrt{3}}. \]
216+72sqrt3
Every angle can be found.
In triangle CAT, use the sine law to find AC
and in triangle BAT, use the sine law to find AB
Area of triangle ABC = (1/2)(AC)(AB)sin 60°
= ...