In triangle ABC, AM, and CN are medians to the sides BC and AB, respectively.
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In triangle ABC, AM, and CN are medians to the sides BC and AB, respectively. If AM and CN are perpendicular to each other, AM = 9 cm and CN = 6 cm, then:
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Answer: A. tan⁻¹(9/7)
Using the centroid property (medians divide in 2:1 ratio) and setting perpendicular medians as coordinate axes with G at the origin, we can place A(0, 6), B(-4, -6), and C(4, 0). The angle BAC between vectors AB and AC is found using tan(θ) = |cross product|/|dot product| = 72/56 = 9/7, giving tan⁻¹(9/7).
Step-by-step Derivation:
- Centroid G divides AM: AG = 6, GM = 3; divides CN: CG = 4, GN = 2.
- Set G at origin with AM on y-axis, CN on x-axis: A(0,6), M(0,-3), C(4,0), N(-2,0).
- M is midpoint of BC: B = 2M - C = (0,-6) - (4,0) = (-4,-6).
- Vectors: AB = (-4,-12), AC = (4,-6).
- tan(∠BAC) = |(-4)(-6) - (-12)(4)| / |(-4)(4) + (-12)(-6)| = |24+48| / |-16+72| = 72/56 = 9/7.
- Answer: tan⁻¹(9/7).