Question

3) TWO
foeces of magnitude 8N and 15N
at point. if
the resultant force is
17N, the
angle between the
forces
has
to be

a.30°
b.45°
C.120°
D.90°​

Answer 1

In the given question:

• Resultant = 17 N
• Magnitude of First force = A = 8 N
• Magnitude of Second force = B = 15 N

We are supposed to find the angle between the two forces.

Apply parallelogram law of vector addition to find out the value of θ

(Refer attachment)

$\pink{\bf{R = \sqrt{ {A}^{2} + {B}^{2} + 2ABcos \theta } }}$

Substitute the given values in the above Equation.

$\dashrightarrow \: \sf{17= \sqrt{ {8}^{2} + {15}^{2}+2(8)(15)cos \theta }}$

$\dashrightarrow \: \sf{17 = \sqrt{64 + 225 + 240 \cos \theta } }$

$\dashrightarrow \: \sf{17 = \sqrt{289 + 240 \cos \theta } }$

Squaring on both sides,

$\dashrightarrow \: \sf{289 = 289 + 240 \cos \theta }$

Cancel 289 on both sides.

$\dashrightarrow \: \sf{240 \cos \theta = 0 }$

$\dashrightarrow \: \sf{cos \theta = 0}$

$\dashrightarrow \: \sf{ \theta = {cos}^{ - 1}(0) }$

$\dashrightarrow \: \bf{ \theta = 90^{ \circ} }$

$\therefore \boxed{ \bf{ \red{Angle \: between \: 2 \: vectors=90 ^{\circ} }}}$