I got this question in my advanced functions class for the trig unit and don't understand it. Please help! A circular dining room at the top of a skyscraper rotates in a counterclockwise direction so diners can see the entire city. A woman sits next to the window ledge and places her purse on the ledge. Eighteen minutes later she realizes that her table has moved but her purse is on the ledge where she left it. The coordinates of her position are (x,y) = (20cos(7.5t)°, 20 sin(7.5t)°), where t is the time in minutes and x and y are in metres. What is the shortest distance she has to walk to retrieve her purse?  

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The woman travels in a circle given by the equation x(t) = 20cos(7.5t), y(t) = 20sin(7.5t). Thus the circle's radius is sqrt(x^2 + y^2) = 20 metres.

Take the purse as θ1 = 0°. Then after 18 min, the woman sits at (-14.1, 14.1) by ( x(t),y(t) ); thus θ2...

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The woman travels in a circle given by the equation x(t) = 20cos(7.5t), y(t) = 20sin(7.5t). Thus the circle's radius is sqrt(x^2 + y^2) = 20 metres.

Take the purse as θ1 = 0°. Then after 18 min, the woman sits at (-14.1, 14.1) by ( x(t),y(t) ); thus θ2 = 135°.

The shortest distance to the purse is the chord from θ1 to θ2 (length d). Draw a line from the center to the chord bisecting the angle to form two identical right triangles. Now,

d/2 = 20Sin(135/2) = 18.48 m

d = 36.96 m

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