Question

0.14
If A=B+C and magnitudes of A, B and C are 5, 4, and 3 units respectively, the angle between A and C
is -
(A) sin- (3/4) (B) cos- (4/5) (C) cos- (3/5) (D) π/2​

Answer 1

Answer:

Option C) cos⁻(3/5)

Explanation:

Given, |A| = 5, |B| = 4 and |C| = 3.

Also, A = Vector sum of B and C.

So we can say that A is the resultant of the vectors B and C.

We know that the resultant (R) of two vectors is given as

$\sf R^2 = {V_1}^2 + {V_2}^2 + 2V_1V_2cos\theta$

Where, $\sf V_1 \ and \ V_2 \ are \ two \ vectors$

So, we can write here

A² = B² + C² + 2BC(cos∅)

→ 5² = 4² + 3² + 2(4)(3)(cos∅)

→ 25 = 16 + 9 + 2(12)(cos∅)

→ 25 = 25 + 24(cos∅)

→ 24(cos∅) = 0

→ cos∅ = 0

→ cos∅ = cos(π/2)

Hence ∅ = π/2 which is the angle between the two vectors.

Now, θ is the angle between B and C. Let the angle between A and C be β. (refer the attachment)

So, in the given attachment, we can write cosβ as 3/5

(cosθ = B/H)

So, β will be cos⁻(3/5)

Answer 2

Solution :-

We are Provided with

A = B + C

And

|A| = 5

|B| = 4

|C| = 3

Now we have got three Vector A , B , C

and A is the Resultant of A and B .

Now as we know

$| \vec{R} | = \sqrt{A^2 + B^2 + 2AB . Cos\theta }$

Now via using it :-

$\rightarrow 5 = \sqrt{ (4)^2 + (3)^2 + 2(4)(3) . cos\theta}$

$\rightarrow 5 = \sqrt{16 + 9 + 12 . cos\theta}$

$\rightarrow 5 = \sqrt{25 + 12.cos\theta}$

By squaring both sides :-

$\rightarrow 25 = 25 + 12.cos\theta$

$\rightarrow 12.cos\theta = 25 - 25$

$\rightarrow 12.cos\theta = 0$

$\rightarrow cos\theta = 0$

Now

∅ = π/2

So angle Between B and C = π/2

Now for further solution refer attached

Answer :- Option (C) Cos⁻ (3/5)