Question #8f4d7

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Question #8f4d7

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Answer 1 · 2016-09-13T12:38:23

${55.8}^{\circ}$ $55.8^@$ , rounded to one decimal place.

Explanation:

Given two vectors
$\to S$ $vecS$ , with $∣ ∣ ∣ \to S ∣ ∣ ∣ = 4$ $|vecS|=4$ and $\to R$ $vecR$ , with $∣ ∣ ∣ \to R ∣ ∣ ∣ = 6$ $|vecR|=6$
Let $θ$ $theta$ be the angle between the two.
By definition of dot product we have
$\to S \cdot \to R = ∣ ∣ ∣ \to S ∣ ∣ ∣ ∣ ∣ ∣ \to R ∣ ∣ ∣ cos θ$ $vecS cdot vecR=|vecS||vecR|cos theta$
Equating to the given value we get
$∣ ∣ ∣ \to S ∣ ∣ ∣ ∣ ∣ ∣ \to R ∣ ∣ ∣ cos θ = 13.5$ $|vecS||vecR|cos theta=13.5$
$\Rightarrow 4 \times 6 cos θ = 13.5$ $=>4xx6cos theta=13.5$
$\Rightarrow cos θ = \frac{13.5}{24} = 0.5625$ $=>cos theta=13.5/24=0.5625$
$\Rightarrow θ = {cos}^{- 1} 0.5625$ $=> theta=cos^-1 0.5625$
$\Rightarrow θ = {55.8}^{\circ}$ $=> theta=55.8^@$ , rounded to one decimal place.

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Question #8f4d7

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