Let ABC be a triangle. Let vectors AB⋅AC= 5✓3 and |AB||AC|=10. Find the area of the triangle ABC. The question comes with a diagram in which the angle CAB is obtuse. Now, I used the definition of dot product to find the angle CAB which is 150 degrees, but I don't know how to find the magnitude of vectors AB and AC (ie. |AB| and |AC|) to find the area of a triangle using the formula A= 1/2 ab sin C. Any help is appreciated.
Area of a triangle (exam question)
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linear-algebra
trigonometry
1 Answers
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We have $cb=|AB||AC|=10$ so the area is $$\frac{1}{2}cb\sin A=5\sqrt{1-\cos^2 A}=5\sqrt{1-\left(\frac{5\sqrt{3}}{10} \right)^2}=\frac{5}{2}.$$
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0Or, more simply, since the OP has already found angle $A,$ we can plug it in directly, without need to detour through Pythagorean identities. – 2017-01-22