We have given triangle $ABC$ where $AB=a$ and we also know that $\angle A=\alpha$ and $\angle B = \beta $ my question is how to find the area thereof triangle? I thing there would be useful formula $[ABC]=\frac{a*|AC|\sin\alpha}{2}$
[Math] Find the area of a triangle knowing one side and two angles.
geometry
Related Solutions
Summing up the area of two triangles we get, $3\sin (\frac{A}{2})+4\sin (\frac{A}{2})=7\sin (\frac{A}{2})=4\sin A$. Thus, $\cos (\frac{A}{2})=\frac78$. Or $\cos A=2(\frac78)^2-1$.
So: $$CF=8\sin A=8\sqrt{1-(2(\frac78)^2-1)^2}=8\sqrt{1-(\frac{17}{32})^2}=8\sqrt{1-\frac{17}{32}}\sqrt{1+\frac{17}{32}}=\frac74\sqrt{15}$$
Use:
In triangle $ABD:$
$$\angle D=90^{\circ},\angle A=30^{\circ} \Rightarrow BD=\frac12AB=10$$
Best Answer
Suggestion: Use the Law of Sines
$$\dfrac {\sin(\angle C)}{a} = \dfrac {\sin \alpha}{|BC|} = \dfrac{\sin \beta}{|AC|}$$