A man is standing at a distance of 60m from one tree and 54m from another tree on a playground. If the area of the triangle formed by the man and two trees is 810m², at what angular position is the man standing with these two trees?
I have been struggling with this question. I am try to solve this question making different scenarios but I am not able to figure out the answer so I am posting it for your help.
In 1st Scenario, I tried making the both trees in same direction but I am stuck in an stage that I am not able to deal with.
In Scenario 2, I supposed that the trees are at different direction but even this seems a bit more realistic, no more contribution it added in my problem solving. I am stuck in finding the sides and the other angles."
That's why I strongly request this math community to solve my question. Before you try this, I want to let you know that this is a District Mathematics Olympiad (DMO) question. It has also provided some options which are as follows:
i.45° ii.60° iii.15° iv.30°
It is known that the area of a triangle is $\frac12 ab\sin(C)$, where $a$ and $b$ are two sides of the triangle and $C$ is the angle between them. We know that this expression is $810$ meters squared, and we can let $a=60$ and $b=54$. So we have $$810=30\times54\sin(C)\implies\frac{810}{30\times54}=\sin(C)\implies\frac12=\sin(C)$$I will let you take it on from here.
Edit: I made it clearer why the area is $\frac12ab\sin(C)$ here