$\mathrm{AB}=10$ and it intersects plane a. $\mathrm{A}$ and $\mathrm {B}$ are $2\mathrm{m}$ and $3\mathrm{m}$ far from the plane and the question is to find angle formed by plane a and $\mathrm{AB}$.
I get it that I can use tangent to find the angle but I only know one side of the both triangles, so how could I approach this problem?
Notice, let the line $AB$ intersect the plane at point $O$ & $AM=2$ & $BN=3$ be the perpendiculars drawn to the plane from the points $A$ & $B$ respectively then $\triangle AMO$ & $\triangle BNO$ are similar right triangles so one should have $$\frac{AM}{BN}=\frac{AO}{OB}\ \ \ \ \ \left(\triangle AMO\sim \triangle BNO\right)$$ $$\frac{2}{3}=\frac{AO}{OB}$$ $$\frac{2}{3}+1=\frac{AO}{OB}+1$$ $$\frac{5}{3}=\frac{AO+OB}{OB}$$ $$\frac{5}{3}=\frac{AB}{OB}$$ $$\frac{5}{3}=\frac{10}{OB}\ \ \ \ \ \ \ \ \left(\text{given}, \ \ \ AB=10\right)$$ $$\implies OB=6$$
If $\theta$ is the angle between plane & the line $AB$ then, in right $\triangle BNO$, $$\sin\theta=\frac{BN}{OB}=\frac{3}{6}=\frac{1}{2}$$ $$\theta=\sin^{-1}\left(\frac{1}{2}\right)=30^\circ$$