Question

Two forces with magnitudes of 150 and 75 pounds act on an object at angles of 30° and 150°, respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.

Answer 1

The problem is modelled in the first picture shown below

To work out the resultant vector, we modelled the vectors 150N and 75N as triangle AOB is shown in the second picture with AB as the resultant vector. 

We use the cosine rule to work out the length AB
$AB^{2}= 75^{2}+ 150^{2}-(2*75*150*cos(60))$
$AB^{2} =28125-11250$
$AB^{2}=16875$
$AB= \sqrt{16875} =130$(nearest whole number)

The third picture shows the full diagram of the vectors

To work out the direction of the resultant vector, we use the sin rule to find the size of angle A and angle B

Angle A
$\frac{130}{sin(60)}= \frac{75}{sin(A)}$
$130sin(A)=75sin(60)$
$sin(A)= \frac{75sin(60)}{130}$
$sin(A)=0.4996300406$
$A= sin^{-1}(0.4996300406)$
$A=30$ (rounded to nearest whole number)

Angle B
$B=180-60-30=90$

Direction is 60° toward negative x-axis

Answer: Magnitude 130N and direction 60° toward negative x-axis