In a triangle ABC , |BC| = 8 , |CA| = 7 , |AB| = 10 , then the projection of the AB on AC is equal to: - Mathematics MCQ Mathematics Mock Test - 9 2026 | ExamTest.live

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In a triangle A BC, ∣ BC ∣ = 8, ∣ C A ∣ = 7, ∣ A B ∣ = 10, then the projection of the A B on A C is equal to:

Solution & Step-by-step Explanation

Let a = BC = 8, b = A C = 7, c = A B = 10 .We need the projection of A B on A C, which is ∣ A B ∣ cos A = c cos A .Using the cosine rule: cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c} cos A = \frac{7 ^{2} + 1 0 ^{2} - 8 ^{2}}{2 ( 7 ) ( 10 )} = \frac{49 + 100 - 64}{140} = \frac{85}{140} = \frac{17}{28} .Projection = c cos A = 10 \times \frac{17}{28} = \frac{170}{28} = \frac{85}{14} .

Try it yourself before checking the explanation above.

In a triangle A BC, ∣ BC ∣ = 8, ∣ C A ∣ = 7, ∣ A B ∣ = 10, then the projection of the A B on A C is equal to:

25/4

85/14

127/20

115/16

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