Question

find the magnitude of the vector sum \\(\vec{a} + \vec{b}\\).
vector a is 4.99 m long in a 5.79° direction.
vector b is 11.5 m long in a 11.3° direction.

Explanation:

Step1: Resolve Vector A into components

To find the x - component of vector \(\vec{A}\), we use the formula \(A_x = A\cos\theta_A\), where \(A = 4.99\space m\) and \(\theta_A=5.79^{\circ}\). So \(A_x=4.99\times\cos(5.79^{\circ})\).
To find the y - component of vector \(\vec{A}\), we use the formula \(A_y = A\sin\theta_A\), where \(A = 4.99\space m\) and \(\theta_A = 5.79^{\circ}\). So \(A_y=4.99\times\sin(5.79^{\circ})\).

Calculating the values:
\(\cos(5.79^{\circ})\approx0.9953\), so \(A_x = 4.99\times0.9953\approx4.97\space m\)
\(\sin(5.79^{\circ})\approx0.1009\), so \(A_y=4.99\times0.1009\approx0.503\space m\)

Step2: Resolve Vector B into components

To find the x - component of vector \(\vec{B}\), we use the formula \(B_x = B\cos\theta_B\), where \(B = 11.5\space m\) and \(\theta_B = 11.3^{\circ}\). So \(B_x=11.5\times\cos(11.3^{\circ})\)
To find the y - component of vector \(\vec{B}\), we use the formula \(B_y = B\sin\theta_B\), where \(B = 11.5\space m\) and \(\theta_B=11.3^{\circ}\). So \(B_y = 11.5\times\sin(11.3^{\circ})\)

Calculating the values:
\(\cos(11.3^{\circ})\approx0.9808\), so \(B_x=11.5\times0.9808\approx11.28\space m\)
\(\sin(11.3^{\circ})\approx0.1963\), so \(B_y=11.5\times0.1963\approx2.26\space m\)

Step3: Find the components of the resultant vector \(\vec{R}=\vec{A}+\vec{B}\)

The x - component of the resultant vector \(R_x=A_x + B_x\)
\(R_x=4.97+ 11.28=16.25\space m\)

The y - component of the resultant vector \(R_y=A_y + B_y\)
\(R_y=0.503 + 2.26=2.763\space m\)

Step4: Calculate the magnitude of the resultant vector

The magnitude of a vector with components \(R_x\) and \(R_y\) is given by \(R=\sqrt{R_x^{2}+R_y^{2}}\)
Substituting \(R_x = 16.25\space m\) and \(R_y=2.763\space m\)
\(R=\sqrt{(16.25)^{2}+(2.763)^{2}}=\sqrt{264.0625 + 7.634}=\sqrt{271.6965}\approx16.48\space m\)

Answer:

The magnitude of \(\vec{A}+\vec{B}\) is approximately \(16.5\space m\) (rounded to three significant figures)