Create a triangle starting with the length of 540 coming out of SF to LA bearing 140.
This looks like it leaves SF in IV quadrant and enters LA in II quadrant. The wind will take the plane more to the east (right). Therefore the pilot must aim to the west (left) in order for the wind to push it to the intended destination of LA. The second side of the triangle should come out of LA from the II quadrant bearing 290. The length of this side is 60t (60 km/h * t = time flying). The third side of the triangle connects the first two. There for it comes out of SF at some unknown angle > 140 and connects with the 60t side. The length of this third leg is 640t. (640 km/h * t = time flying). The angle between the 540 side and the 60t side is 30. This is found because the 540 side enters LA in 2nd quadrant as 130 angle . The 60t side enters LA in 2nd quadrant as 160 angle. Using law of sin's: sin 30/640t = sin x/60t. The t's cancel and you are left with sin x = 3/64. When solving for the angle x you get x = 2.6867 degrees. Adding this to the bearing of 140, the compass bearing should be 142.6867 or 142.7 degrees. To find the value of t, you use the law of sin's to get sin 30/640t = (sin (180 - 30 - 2.6867))/540. Solving for t gives you: t = .7811. This is in hours. To convert to minutes multiply by 60 to get t = 46.87 minutes. Add this to the 2pm departure time to get 2:47 pm arrival time.
The phytagorean theorem states a^2 + b^2 = c^2
And so your answer is C. (UV)^2+(VW)^2=(UW)^2, because 81+144=225
Since they are squares, the length of one sides square is equal to the area. UV, VW, and UW are all sides on a square. If you square them, you get their area, and then if you add the areas it will prove the phytagorean theorem correct.
Answer:
8 years older then your younger sister
