An aircraft flies a certain distance on a bearing of 135 deg

An aircraft flies a certain distance on a bearing of 135 degree and then twice the distance on a bearing of 225 degree. its distance from the starting point is then 350km . Find the length of the first part of the journey.

Solution

LET THE PLANE START AT A AND GO TO B WHICH IS 135 DEGREE POITION.
LET AB=X
FROM B LET IT GO TO C IN BC DIRECTION WHICH IS 225 DEGREES.THAT IS ANGLE ABC IS 225-135=90 DEGREES.
BC=2X AS GIVEN.
SO IN RIGHT ANGLE TRIANGLE
AC^2=AB^2+BC^2
350^2=X^2+(2X)^2=X^2+4X^2=5X^2
X^2=350*350/5=70*70*5
X=70*SQRT(5) IS THE LENGTH OF FIRST PART OF JOURNEY.