Trigonometry question!?

Question

Triangle ABC has a right angle at A.The hypotenuse BC is trisected at M and N so that BM=MN=NC.If AM=x,AN=y;find the length of MN in terms of x and y.

Answer 1

Let the sides opposite the angles be a, b, and c.
Then

a² = b² + c²
MN = a/3
MN² = a²/9

Drop a perpedicular from N to b and call it D.
Drop a perpedicular from N to c and call it E.
Now we have triangle ADN with a right angle at D.
AD = 2b/3
ND = c/3
y² = AD² + ND² = (2b/3)² + (c/3)² = 4b²/9 + c²/9

Drop a perpedicular from M to b and call it F.
Drop a perpedicular from M to c and call it G.
Now we have triangle AFM with a right angle at F.
AF = b/3
MG = 2c/3
x² = AF² + MG² = (b/3)² + (2c/3)² = b²/9 + 4c²/9

Now add equations.

x² + y² = {b²/9 + 4c²/9} + {4b²/9 + c²/9} = 5b²/9 + 5c²/9
x² + y² = (5/9)(b² + c²) = 5a²/9 = 5MN²
MN² = (x² + y²)/5
MN = √[(x² + y²)/5]

Answer 2

sounds like this is your HW... go do it! open & read your book!