A sailboat race course marks out a roughly triangular shape from the starting li

Question

A sailboat race course marks out a roughly triangular shape from the starting line, around 2 bouys and back to the starting point. The first leg of the race is 40 km at an angle of 20 north of east. The second leg is 82 km at an angle of 23 south of east. What is the length of the third leg of the race, and the angle?

***The Third leg of the race is 110 km, and the angle is -9.2. ****

I can't seem to visualize where to begin/ I do not know what steps to use. I have seen the other answers to this similar problem on here, please do not answer this question in a similar manner, because I do not understand how to solve for the 3rd leg of the race in detail, so the rest of the answer is useless.

Explanation / Answer

Given The length made by first leg d1= 40 km at 1= 20^0 Length made by second leg d2 = 82 km at 2 =23^0 d1 = 40 km cos20 (i) + 40km sin 20 (j) d1 = 37.5 (i) + 13.6 (j) d2 = 82 km cos 23 (i) + 82 km sin 23 (-j) d2 = 75.4 (i) + 32 (-j) The length made by the third leg in the race is, d = d1 + d2    = (37.5 +75.4) i + (13.6 -32) j d =  112.9 (i) + 18.4 (-j) d = 114 km Angle , = tan^-1(-18.4 /112.9)              = -9.20

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