From the point M lying inside the triangle ABC, perpendiculars are drawn to the sides BC, AC, AB, the lengths of which are k, l and m, respectively. Find the area of a triangle ABC if ∠CAB = α and ∠ABC = β. α = π / 4, β = 2π / 3, k = 5, l = 3, m = 4. If the answer is not an integer, round it to the nearest integer.
I built circles on the segments MA, MB and MC as on diameters ... they will pass through the bases of the perpendiculars lowered to the sides ...
it turns out the task is to find the area of an inscribed quadrilateral along two sides and the angle between them, if the other two angles adjacent to these sides are straight lines ...
maybe there is some ready-made formula for the solution, but I don't know it ... but there is enough data
Let [] denote areas. Consider the quadrilateral AYMZ
$$[AYMZ] = \frac12 m^2 \cot a_1 + \frac12 l^2 \cot a_2$$ $$l\sin a_1 = m \sin a_1, \>\>\>\>\> a_1+a_2 = \alpha$$
Eliminate $a_1$ and $a_2$ to get $$[AYMZ]=\frac12(m^2+l^2)\cot \alpha + ml\csc \alpha$$
Similarly, \begin{align} [BZMX] &=\frac12(k^2+m^2)\cot \beta+ km\csc \beta\\ [CXMY] &=\frac12(l^2+k^2)\cot \gamma+ lk\csc \gamma\\ \end{align} where $\gamma = \pi - (\alpha+\beta)$. Thus, the area of the triangle ABC is \begin{align} [ABC] &= [AYMZ] + [BZMX] + [CXMY] \\ & = \frac12[ (m^2+l^2)\cot \alpha+(k^2+m^2)\cot \beta+ (l^2+k^2)\cot \gamma]\\ &\hspace{1cm}+ ml\csc \alpha+ km\csc \beta + lk\csc \gamma \end{align}