# Addition of a magnitude of a vector [closed]

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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## closed as too localized by Austin Mohr, William, Noah Snyder, Ｊ. Ｍ., NorbertOct 7 '12 at 13:58

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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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so i my answer is wrong? –  JackyBoi Sep 12 '12 at 13:56
@JackyBoi: Your answer is right. Note that in my post, I wrote that the square of the diagonal is about $2156.4204$. If you take the square root you will get same answer as the one you got. I was just giving another way of finding the answer. But since it caused confusion, I will write down the square root. –  André Nicolas Sep 12 '12 at 14:04
${\bf a}=\langle 27\cos(140^\circ),27\sin(140^\circ)\rangle \approx \langle -20.683, 17.355 \rangle$ 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.