Can you help this maths question on Bearings?

Simon travels 1414 m due North from the school then heads East for 1414 m. He finally changes the course and travels 5000m sw. His friend, Tom, is some distance from school and on the bearing SE from his school.

How far is Simon from the school finally

b) What is Simon's final bearing from school

c) Suppose Tom is due East of Simon, find the distance from Tom to the school

Can you help this maths question on Bearings?
Since Simon travels due north for the same distance as due east, when he goes southwest, he passes through the school. The plot of his path is a right triangle with the two sides = 1414 m. The distance that he travels to get back to the school is the hypotenuse of this right triangle.



Therefore Simon is exactly southwest of the school...%26lt;%26lt;%26lt;...first answer



Let c = the length of the hypotenuse in meters



The Pythagorean Theorem states that:



c^2 = sqrt[(1414)^2 + (1414)^2] = sqrt(1,999,396 + 1,999,396) =



sqrt(3,998,792) = 1,999.698 m



(Note: Since the square root of 2 is usually approximated by 1.414, I suspect that whoever wrote this question wanted the length of 1414 to be approximated by 1000*sqrt(2), but that is only an approximation so I'll use the numbers that I computed.)



Since Simon traveled a total of 5000 m and 1999.698 was the distance that he traveled to the school, the distance that he traveled beyond the school is:



5000 - 1999.698 = 3000.302



Since Tom is southeast of the school and east of Simon, this is another right triangle with the right angle based at the school and the distance from Tom to the school is:



3000.302 m....%26lt;%26lt;%26lt;...Second answer

school nurse needing advice