Apilot flies in a straight path for 1 hour and 30 min. she then makes a course correction, heading 10 degrees to the right of her original course, and flies 2 hours in the new direction. if she maintains a constant speed of 640 miles per hour, how far is she from her starting position?

The first leg of the flight is (!.5 hr)(680 mph) = 1020 miles long.  Let's call this side b of our triangle.The second leg of the flight is (2 hr)(680 mph) = 1360 miles long. Let's call this side c of our triangle. We can find the length of side a of the triangle from the law of cosines.  The angle A between the two sides whose length we know is 170°. (180° - 10° = 170°). a2 = b2 + c2 - (2)*(a)*(b)*cos(A) a2 = (1020 miles)2 + (1360 miles)2 - (2)*(1020 miles)(1360 miles)*cos(170°) a2 = 1040400 miles2 + 1849600 miles2 - (2774400 miles2)*cos(170°) cos(170°) = -0.9848 a2 = 2890000 miles2 + 2732251 miles2 = 5622250.6 miles2 a = (5622250.6 miles2)½ = 2371 miles

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2.4+3.2t. you get the answer by adding the ones with variables and ones without