Question

Angle between two vectors of magnitudes 12 and 18 units when their resultant is 24 units is

Answer 1

Hi..

R² = A² + B² + 2ABcos(θ)
24² = 12² + 18² + 2*12*18cos(θ)

We can do all the arithmetic or spot that 6² cancels out, giving:
4² = 2² + 3² + 2*2*3cos(θ)
16 = 13 + 12cos(θ)
cos(θ) = 3/12 = 0.25
θ = cos⁻¹0.25 = 75.5º..

Hope this helps u!!

Answer 2

Answer:

75°52'

Explanation:

Given:

Magnitude of first vector$(\vec{A})=12$

Magnitude of second vector$(\vec{B})=18$

and resultant of the given vectors $(\vec{R})=24.$

Find

Resultant vector

Solution

Magnitude of first vector$(\vec{A})=12$

Magnitude of second vector $(\vec{B})=18$

and resultant of the given vectors $(\vec{R})=24$

We know that resultant vector

$|\vec{R}|=24=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}=\sqrt{(12)^{2}+(18)^{2}+2 \times 12 \times 18 \cos }$

or $(24)^{2}=144+324+432 \cos \theta$

$\Rightarrow 432 \cos \theta=108$

$\Rightarrow \cos \theta=\frac{108}{432}=0 \times 25$

or $\theta=\cos ^{-1} 0 \times 25=75^{\circ} 52^{\prime}$

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