Question

find the magnitude of the vector sum \\(vec{a} + vec{b}\\).
vector a is 136 m long in a \\(62^circ\\) direction.
vector b is 96.1 m long in a \\(31^circ\\) direction.

Explanation:

Step1: Resolve vectors into components

For vector \(\vec{A}\) (magnitude \(A = 136\) m, angle \(\theta_A = 62^\circ\)):
\(A_x = A\cos\theta_A = 136\cos62^\circ\)
\(A_y = A\sin\theta_A = 136\sin62^\circ\)

For vector \(\vec{B}\) (magnitude \(B = 96.1\) m, angle \(\theta_B = 31^\circ\)):
\(B_x = B\cos\theta_B = 96.1\cos31^\circ\)
\(B_y = B\sin\theta_B = 96.1\sin31^\circ\)

Calculate the components:
\(A_x = 136\cos62^\circ \approx 136\times0.4695 \approx 63.85\) m
\(A_y = 136\sin62^\circ \approx 136\times0.8829 \approx 120.07\) m
\(B_x = 96.1\cos31^\circ \approx 96.1\times0.8572 \approx 82.37\) m
\(B_y = 96.1\sin31^\circ \approx 96.1\times0.5150 \approx 49.49\) m

Step2: Find the components of the resultant vector

Resultant \(x\)-component: \(R_x = A_x + B_x = 63.85 + 82.37 = 146.22\) m
Resultant \(y\)-component: \(R_y = A_y + B_y = 120.07 + 49.49 = 169.56\) m

Step3: Calculate the magnitude of the resultant vector

Use the Pythagorean theorem: \(R = \sqrt{R_x^2 + R_y^2}\)
\(R = \sqrt{(146.22)^2 + (169.56)^2}\)
\(R = \sqrt{21380.29 + 28750.59}\)
\(R = \sqrt{50130.88} \approx 223.9\) m

Answer:

The magnitude of \(\vec{A} + \vec{B}\) is approximately \(\boxed{224}\) m (rounded to a reasonable precision).