Question:
An engineer standing at point \(P\) wishes to determine the width of a rectangular pond. She finds distance to westernmost point \(A\) to be \(60\) m and distance to northernmost point \(B\) to be \(80\) m. If angle \(APB=60^\circ\), find \(AB\).
An engineer standing at point \(P\) wishes to determine the width of a rectangular pond. She finds distance to westernmost point \(A\) to be \(60\) m and distance to northernmost point \(B\) to be \(80\) m. If angle \(APB=60^\circ\), find \(AB\).
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Whenever two sides and included angle are given, use the cosine rule directly.
Updated On: Jun 11, 2026
- \(13\sqrt{20}\)
- \(13\sqrt{10}\)
- \(20\sqrt{13}\)
- \(10\sqrt{13}\)
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The Correct Option is C
Solution and Explanation
Step 1: Use the cosine rule in triangle \(APB\).
Given \[ PA=60, \qquad PB=80, \qquad \angle APB=60^\circ. \] By cosine rule, \[ AB^2 = PA^2+PB^2 -2(PA)(PB)\cos60^\circ. \] \[ = 60^2+80^2 -2(60)(80)\left(\frac12\right). \] \[ = 3600+6400-4800. \] \[ = 5200. \]
Step 2: Find \(AB\).
\[ AB = \sqrt{5200} = 20\sqrt{13}. \] Therefore, \[ \boxed{20\sqrt{13}}. \]
Given \[ PA=60, \qquad PB=80, \qquad \angle APB=60^\circ. \] By cosine rule, \[ AB^2 = PA^2+PB^2 -2(PA)(PB)\cos60^\circ. \] \[ = 60^2+80^2 -2(60)(80)\left(\frac12\right). \] \[ = 3600+6400-4800. \] \[ = 5200. \]
Step 2: Find \(AB\).
\[ AB = \sqrt{5200} = 20\sqrt{13}. \] Therefore, \[ \boxed{20\sqrt{13}}. \]
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