First slide

Vectors

Question

Three forces 9 N, 12 N and 15 N acting at a point are in equilibrium. The angle between 9 N and 15 N is

Moderate

A

$\cos^{- 1} (4 / 5)$
B

$\cos^{- 1} (3 / 5)$
C

$π - \cos^{- 1} (3 / 5)$
D

$π - \cos^{- 1} (4 / 5)$

Solution

${We can see that (9)}^{2} + (12)^{2} = 15^{2} So, forces 9 and 12 are perpendicular to each other.$
$\begin{array}{l} \tan α = \frac{12}{9} = \frac{4}{3} \\ \sin α = 4 / 5 \\ Angle between 9 N and 15 N is \\ π - α = π - \cos^{- 1} (3 / 5) \end{array}$

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