ABC is a triangle, right angled at A.

Question

$ABC$ is a triangle, right angled at $A$. The resultant of the forces acting along $AB$, $AC$ with magnitudes $\dfrac1{AB}$ and $\dfrac1{AC}$ respectively is the force along $AD$, where $D$ is the foot of the perpendicular from $A$ onto $BC$. The magnitude of the resultant is:

Accepted Answer

Resultant $=\dfrac1{AB}\hat i+\dfrac1{AC}\hat j$ (unit vectors along $AB$, $AC$), so magnitude $=\sqrt{\dfrac1{AB^2}+\dfrac1{AC^2}}=\dfrac{\sqrt{AB^2+AC^2}}{AB\cdot AC}=\dfrac{BC}{AB\cdot AC}$ (using $AB^2+AC^2=BC^2$).

Since $	riangle ABC\sim	riangle DBA$, $\dfrac{BC}{AB}=\dfrac{AC}{AD} \implies \dfrac{BC}{AB\cdot AC}=\dfrac1{AD}$

Solutions · 1