If the area of a triangle is 9 √ 2 and two of the sides have length 6 and 3, find the angle between these two sides.

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Okay, so we have the following situation: aside of length: 3, aside of length 6. So let's do this way, 6, actually, 3 and 6, and we need to find the length of this angle of this okay. We need to find this angle here, theta. Okay, now we can think of these 2 sides as vectors, so this 1 is a vector v. This 1 is a vector point now we know that the area of this triangle- the area of this triangle- is a equals 9 square root of 2 point now. We also know that this area is equal to the length of the vector v, multiplied by the length of the vector multiplied by sine of theta and finally multiplied by 2 point. So what do we have? We have 9 square root. 2, equals 18 actually 3036, because this 1 multiplied by the son multiplied by sine of theta. So here what do we get? We get sine of theta equals square root of 2 over 4, which implies theta equals arc sine or square to 2. Over 4 point so are done.

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