Question

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?

Answer 1

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°
= ...

Answer 2

i got AC=24 and AB= 32.78. This gave me 340.7 which was wrong. What should i do?

Answer 3

I got the same results, and I stand by them

Answer 4

216+72sqrt3

Answer 5

##### 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
\[[ABC]= \frac{1}{2}(AB)(CH)= \frac{1}{2}(12 + 12\sqrt{3})(12\sqrt{3}) =\boxed{ 216 + 72\sqrt{3}}. \]

Answer 6

Lol you literally copied the question from AoPS... this is against the honor code but whatever... u can ask stuff on the message board.

Answer 7

and you also copied the answer bro... if ya gonna plagiarize at least do it well