Question

Magnitudes of Ā,B,C are 3,4,5. Are form a triangle, when taken in clock wise order perpendicular at B then
(A) Angle between A and B is 90°
(B) Angle between A and C = sin-1 (4/3)
(C) Angle between B and C = sin-1(3/5)
(D) Angle betweenĀ and C = sin-1(3/4)​

Answer 1

A

Angle between A and B is 90

Answer 2

Given:

$\underset{A}{\rightarrow}$,$\underset{B}{\rightarrow}$,$\underset{C}{\rightarrow}$ forms a triangle and their magnitudes are 3,4,5 respectively.

To find:

Angle

Solution:

For calculated the angle between $\underset{A}{\rightarrow}$ and $\underset{B}{\rightarrow}$ .

Suppose  $\theta$ be the angle between $\underset{A}{\rightarrow}$ and $\underset{B}{\rightarrow}$ then,

$\underset{A}{\rightarrow}$ + $\underset{B}{\rightarrow}$ =  $\underset{C}{\rightarrow}$

 by applying square both sides in the above equation, we get

$(\vec{A} + \vec{B} )^{2} = (\vec{C})^{2}$

$\left | \vec{A} \right |^{2} + \left | \vec{B} \right |^{2}+ 2.\vec{A}.\vec{B} = \left | \vec{C} \right |^{2}$

$\left | \vec{A} \right |^{2} + \left | \vec{B} \right |^{2}+ 2\left | \vec{A} \right |\left | \vec{B} \right |cos\theta = \left | \vec{C} \right |^{2}$

by putting all the values in above equation we get,

$\left {3} ^{2} + \left {4} ^{2}+ 2\times 3 \times 4 cos\theta = \left {5} ^{2}$

$cos\theta = 0$

$\theta =90^{\circ}$

Therefore, angle between $\vec{A}$ and $\vec{B}$ is 90°.

So, Option(A) is correct.