Tammy leaves the office, drives 36 km due north.Then she turns onto a second highway and continues in direction 45 degree north of east for 50 km. What is her total displacement from the office? (give magnitude and direction)

Sketching the displacement vectors roughly to scale is a big help in seeing what needs to be done.

Draw the first arrow vertically on your page, with the arrow pointing up, and label it 35 km.
Start the next vector from the tip of that arrow, heading 45 ° clockwise from North; make it longer than the first vector; label it 50 km.
Join the tail of the first vector to the tip of the second vector, and that represents your resultant displacement (it points upward/rightward).

You should now see that the angle between the 35 km N and 50 km NE vectors is 45 ° + 90 ° = 135 °
Thus, you can use the cos law to find the magnitude of the resultant displacement.
Once you have this number, label it on your diagram.
You now need the direction of this vector, which will be the angle between it and the 35 km N vector. The sin law will help you here. The angle you find can be reported as N ___ ° E.

I agree, law of cosines is the way to go.

To find Tammy's total displacement, we can break down her movements into two separate displacements: one due north and one at a 45-degree angle north of east.

For the first displacement, Tammy drives 36 km due north. Since this is a straight line, the magnitude of her displacement in the north direction is simply 36 km.

For the second displacement, Tammy travels 50 km at a 45-degree angle north of east. To find the north and east components of this displacement, we can use basic trigonometry.

The north component would be the length of the side adjacent to the 45-degree angle, and the east component would be the length of the side opposite the 45-degree angle.

Using trigonometry, we can determine that the north component (x) is given by: x = 50 km * cos(45 degrees) ≈ 50 km * 0.7071 ≈ 35.36 km

Similarly, the east component (y) is given by: y = 50 km * sin(45 degrees) ≈ 50 km * 0.7071 ≈ 35.36 km

Now we can calculate the total displacement by summing the north displacement components and the east displacement components:

Total north displacement = 36 km + 35.36 km = 71.36 km
Total east displacement = 35.36 km

To find the magnitude of the total displacement, we can use the Pythagorean theorem:

Magnitude = √(Total north displacement)^2 + (Total east displacement)^2
= √(71.36 km)^2 + (35.36 km)^2
≈ √(5099.1296 km^2 + 1250.0096 km^2)
≈ √6349.1392 km^2
≈ 79.7 km (rounded to one decimal place)

Therefore, Tammy's total displacement from the office is approximately 79.7 km.

To find the direction of the displacement, we can use trigonometry again. The direction would be the angle between the positive x-axis (east) and the displacement vector. Using inverse trigonometric functions, we find:

Angle = arctan(Total east displacement / Total north displacement)
= arctan(35.36 km / 71.36 km)
≈ arctan(0.495)
≈ 26.9 degrees (rounded to one decimal place)

So, Tammy's total displacement from the office is approximately 79.7 km in a direction of 26.9 degrees north of east.