Question:
In a triangle ABC with usual notations if \(|\overline{\mathrm{BC}}|=8,|\overline{\mathrm{CA}}|=7,|\overline{\mathrm{AB}}|=10\) then the projection of \(\overline{\mathrm{AB}}\) on \(\overline{\mathrm{AC}}\) is
In a triangle ABC with usual notations if \(|\overline{\mathrm{BC}}|=8,|\overline{\mathrm{CA}}|=7,|\overline{\mathrm{AB}}|=10\) then the projection of \(\overline{\mathrm{AB}}\) on \(\overline{\mathrm{AC}}\) is
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The projection of AB on AC is \(|AB|\cos A\), and \(\cos A\) comes from the cosine rule.
Updated On: Nov 13, 2026
- \(\frac{14}{85}\) units
- \(\frac{1}{85}\) units
- \(\frac{85}{14}\) units
- \(\frac{7}{85}\) units
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
The projection of \(\overline{AB}\) on \(\overline{AC}\) is \(\dfrac{\overline{AB}\cdot\overline{AC}}{|\overline{AC}|} = |AB|\cos A\).
Step 2: Find cos A
Take \(a = 8\), \(b = 7\), \(c = 10\).
\(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{49 + 100 - 64}{140} = \dfrac{85}{140}\)
Step 3: Calculation
Projection \(= 10 \times \dfrac{85}{140} = \dfrac{85}{14}\) units.
Final Answer:
The projection is \(\dfrac{85}{14}\) units. \[ \boxed{\text{(C) }\dfrac{85}{14}\text{ units}} \]
The projection of \(\overline{AB}\) on \(\overline{AC}\) is \(\dfrac{\overline{AB}\cdot\overline{AC}}{|\overline{AC}|} = |AB|\cos A\).
Step 2: Find cos A
Take \(a = 8\), \(b = 7\), \(c = 10\).
\(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{49 + 100 - 64}{140} = \dfrac{85}{140}\)
Step 3: Calculation
Projection \(= 10 \times \dfrac{85}{140} = \dfrac{85}{14}\) units.
Final Answer:
The projection is \(\dfrac{85}{14}\) units. \[ \boxed{\text{(C) }\dfrac{85}{14}\text{ units}} \]
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