# How i show that angle between vectors 5i-4j , 2i+3j is an obtuse?

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### 1 Answer

You need to remember what is the measure of an obtuse angle, hence, all angles whose measures are larger than `90^o` , are called obtuse angles.

You should use the dot product to find the angle between the given vectors to state if the angle is obtuse or not.

`(5bari - 4barj)(2bari + 3barj) = |5bari - 4barj||2bari + 3barj|*cos alpha`

`alpha` represents the angle comprised by the given vectors.

Evaluating the dot product yields:

`5*2 - 4*3 = sqrt(5^2 + 4^2)*sqrt(2^2 + 3^2)*cos alpha`

`10 - 12 = sqrt((25 + 16)(4 + 9))*cos alpha`

`-2 = sqrt(533)*cos alpha =gt cos alpha = -2/sqrt(533) =gt alpha ~~ 95^o`

Notice that the value of cosine function is negative, hence, the cosine function is negative in quadrants 2 and 3, hence `alpha` > `90^o` since in the quadrants 2 and 3, the measures of angles start from `90^o` and go to `270^o` .

**Hence, evaluating the angle between the given vectors, yields that this angle is obtus since `alpha = 95^o` .**

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