A student walk 60m on a bearing of 028 degree and then 180m due east. How is she from her starting point, correct to the nearest whole number?

she is 215m from her starting point

Step-by-step explanation:

The diagram representing the situation is shown in the attached photo. Triangle ABC is formed with AC representing her distance from her starting point.

Angle B = 28 + 90 = 118°

To determine AC, we would apply the law of Cosines which is expressed as

a² = b² + c² - 2abCosA

Where a,b and c are the length of each side of the triangle and B is the angle corresponding to b(AC). Likening the expression to the given triangle, it becomes

AC² = AB² + BC² - 2(AB × BC)CosAC

AC² = 60² + 180² - 2(60 × 180)Cos118

AC² = 3600 + 32400 - 21600Cos118

AC² = 36000 - 21600(-0.46947156)

AC² = 36000 + 10140.59

AC² = 46140.59

AC = √46140.59

AC = 215 m

x = 12

y = 15

step-by-step explanation: