Question

Two force of magnitude 8N and 15 N respectively act at a point. if the resultant force is 17 N , the angle between the forces has to be.​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Let the two vectors be inclined at θ.

Let one Force be ${ ( \vec{A}) = 8 \: N}$

and,

One Force be ${ ( \vec{B}) = 15 \: N}$

$\huge{\bold{\underline{Explanation:-}}}$

The resultant is ${( \vec{R}) = 17 \: N}$

Applying vector formula,

$\large{\bold{R = \sqrt{A^2 + B^2 + 2AB \cos \theta}}}$

Substituting the values,

$\large{17 = \sqrt{(8)^2+ (15)^2 + 2 \times 8 \times 15 \cos \theta}}$

Squaring on both sides,

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

It becomes,

$\large{(17)^2 = (8)^2 + (15)^2 + 2 \times 8 \times 15 \cos \theta}$

Now,

$\large{289 = 64 + 225 + 2 \times 120 \cos \theta}$

$\large{289 = 289 + 240 \cos \theta}$

$\large{289 - 289 = 240 \cos \theta}$

$\large{0 = 240 \cos \theta}$

It becomes,

$\large{ \cos \theta = 0}$

$\large{ \cos \theta = \cos 90 \degree}$

$\huge{\boxed{\boxed{ \theta = 90 \degree}}}$

So, the angle between the two vectors is 90°.

Answer 2

Answer:

$\\ r = \sqrt{ {a}^{2} + {b}^{2} + 2abcos \cos(theta) }$

A=8, B=15, R= 17

17^2=8^2+15^2+2×8×15×costheta

289=64+225+240costheta

= 289=289+24 costheta

24 cos theta =0

cos theta =0=theta=90°