ABC is a triangle, right-angled at A . The resultant of the forces acting along AB and AC with magnitudes 1/AB and 1/AC respectively is a force along - Mathematics MCQ AIEEE 2006 2026 | ExamTest.live

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A BC is a triangle, right-angled at A . The resultant of the forces acting along A B and A C with magnitudes \frac{1}{A B} and \frac{1}{A C} respectively is a force along A D, where D is the foot of the perpendicular from A onto BC . The magnitude of the resultant is:

Solution & Step-by-step Explanation

Let the forces be F_{1} along A B and F_{2} along A C .Since the triangle is right-angled at A, A B and A C are perpendicular.The magnitude of the resultant force R is given by: R = (\frac{1}{A B})^{2} + (\frac{1}{A C})^{2} R = \frac{A B ^{2} + A C ^{2}}{A B ^{2} \cdot A C ^{2}} = \frac{A B ^{2} + A C ^{2}}{A B \cdot A C} In △ A BC, by Pythagoras theorem, BC = A B^{2} + A C^{2} .Also, the area of △ A BC = \frac{1}{2} \cdot A B \cdot A C = \frac{1}{2} \cdot BC \cdot A D .Thus, A B \cdot A C = BC \cdot A D .Substituting these into the expression for R : R = \frac{BC}{BC \cdot A D} = \frac{1}{A D}

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A BC is a triangle, right-angled at A . The resultant of the forces acting along A B and A C with magnitudes \frac{1}{A B} and \frac{1}{A C} respectively is a force along A D, where D is the foot of the perpendicular from A onto BC . The magnitude of the resultant is:

\frac{A B ^{2} + A C ^{2}}{( A B ) ^{2} ( A C ) ^{2}}

\frac{( A B ) ( A C )}{A B + A C}

\frac{1}{A B} + \frac{1}{A C}

\frac{1}{A D}

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