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The magnitude of vector a is 27, and the magnitude of vector b is 30. The direction, in degrees, for the vector a is 140, and the direction, in degrees, for the vector b is 69. What is the magnitude, correct to 2 decimal places, of the vector r, where r = a + b? i changed in to component form and then i resolved and got this answer 46.44 is that correct? |
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Your procedure is correct, and finding the components of a vector in specified directions (not necessarily the positive $x$ and $y$ directions) is very important. There are a number of more geometric approaches. Draw a picture. Our sum is a diagonal vector of a certain parallelogram. The angle between the vectors is $71^\circ$, so the diagonal is the third side of a triangle two of whose sides are $27$ and $30$, and where the angle between the sides is $180^\circ-71^\circ=109^\circ$. So by the Cosine Law, the square of the diagonal is equal to $$27^2+30^2-(2)(27)(30)\cos(109^{\circ}).$$ That gives diagonal $\approx 46.437274$. Now round to two decimal places. |
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${\bf a}=\langle 27\cos(140^\circ),27\sin(140^\circ)\rangle \approx \langle -20.683, 17.355 \rangle$ and ${\bf b}=\langle 30\cos(69^\circ),30\sin(69^\circ)\rangle \approx \langle 10.751,28.007 \rangle$. ${\bf a}+{\bf b}\approx\langle -9.932,45.362 \rangle$ so $\|{\bf a}+{\bf b}\| \approx \sqrt{(-9.932)^2+(45.362)^2}\approx 46.44$. So looks good to me. |
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