A friend walks away from you a distance of 545 m, and then turns (as if on a dim

Question

A friend walks away from you a distance of 545 m, and then turns (as if on a dime) an unknown angle, and walks an additional 191 m in the new direction. You use a laser range-finder to find out that his final distance from you is 432 m. What is the angle between his initial departure direction and the direction to his final location? (Enter only the magnitude.)


18.36 is what i got for the first part using trig idenitys
law of cos



Through what angle did he turn? (Enter only the magnitude.)
this is what I want to learn know please show steps

Explanation / Answer

Sorry, the system didn't allow me to draw the diagram
Basically, you have a triangle with 3 sides, of A = 545, B = 191, and C = 432. The angle between A and C is 18.36°, which you correctly found. Then we need the angle between A and B, which we will call ?. We use the law of sines to figure it out.
According to this law, the ratio of a side of the triangle and the sine of the opposite angle is the same for all combinations of sides and opposite angles. So we have
sin(?)/ C = sin(18.36)/B
sin(?) = C* sin(18.36)/B sin(?) = 432*sin(18.36) / 191 sin(?) = 432*sin(18.36) / 191 sin(?) = 432* 0.3149/ 191 sin(?) = 432* 0.3149/ 191 sin(?) = 0.712430 ? = arcsin(0.712430) ? = 45.432°
sin(?) = 0.712430 ? = arcsin(0.712430) ? = 45.432°

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