areacontest-matheuclidean-geometrygeometry
In the figure below, $ABC$ is a triangle such that $\angle BAC = 150^\circ$ , $BC = 74 cm$ and point $D$ is on $BC$ such that $BD=14cm$.If $\angle ADB=60^\circ$,then what is the area, in $cm^2$, of triangle $ABC$?
I have no idea how to solve this, I think the height can somehow be calculated, but I dont know how, help aswell as solutions would be appreciated
Taken from the 2019 IMC
When I see that $\angle CAB$ is partitioned as $24^\circ$ and $18^\circ$, what angle is left to make an $60^\circ$ angle (which is for making an equilateral triangle, which gives us more equal sides)? After doing that, we see that $\angle ECB = 36^\circ$ and this makes $\angle ABE = 30^\circ$. Now, notice that $\Delta ADB$ is congruent to $\Delta AEB$. So we have $|AD| = |AE| = |AC|$. So the answer is $78^\circ$.
Well, afterall this is an IMO 2001 geometry problem and you can find two solutions, along with a solutions of the other problems from that year, here:
Best Answer
Construct the isosceles triangle ADE. Then, $\angle$AEC = 180 - $\frac 12\angle$ADB = 150. Thus, ACE and ACB are similar triangles with
$$AC^2=BC(DC - DE)\tag{1}$$
Apply the cosine law to the triangle ADC,
$$AC^2=AD^2+DC^2-AD\cdot DC\cos120\tag{2}$$
Let AD = DE = $a$. Eliminate AC in (1) and (2), and substitute other segments with $a$ and the given lengths,
$$a^2+60^2+60a=74(60-a)$$
which yields $a=6$. The area of the triangle is, then,
$$\frac 12 BC\cdot a\sin60= 111\sqrt 3$$