Quick Question

Find the area of triangle ACE.

Hello Qube73!

$\overline{AE}=\sqrt{22^2+10^2}=24.166\\ \angle CAE=\ atan \frac{10}{22}=24.444^\circ\\ \overline{CE}=\sqrt{\overline{AE}^2+30^2-2\cdot 24.166\cdot 30\cdot cos(\angle\ CAE)}\\ \overline{CE}=\sqrt{24.166^2+30^2-2\cdot 24.166\cdot 30\cdot cos\ 24.444^\circ}\\ \overline{CE}=12.806\ \color{Unnecessary !}$

$A_{ACE}=\frac{1}{2}\cdot 30\cdot \overline{AE}\cdot sin\ \angle CAE\\ A_{ACE}=\frac{1}{2}\cdot 30\cdot 24.166\cdot sin\ 24.444^\circ\\ \color{blue}A_{ACE}=150$

  !

Thanks Qube73 for acknowledging Asinus's solution.

I have done it independantly of asinus and got the same answer.

So maybe you would like to explain why you believe our answer is incorrect. 

Hey y'all, that wasn't me... that question was posted at almost 1 AM my time, I wasn't even awake at that point. Asinus' solution was correct, so whoever said that it was incorrect, was

(A - Not me.

(B - Doing a different question.

I'm sorry for any incovience this may have caused.