Question

find the indicated lengths: given KL parallel NM and LM=45, the measure of angle M=50 degrees, and KN perpindicular NM. and NL perpindicular LM find KN and KL

Answer 1

Put a coordinate system with L at the origin and LM on the x-axis. Then
- L = (0,0)
- M = (45,0) (since LM = 45)
- because NL ⟂ LM and passes through L, N = (0, n) for some n > 0.

Now KL ∥ NM, so the slope of KL equals the slope of NM = (0 − n)/(45 − 0) = −n/45. Thus KL is the line y = −(n/45)x.
KN ⟂ NM, so KN has slope 45/n and passes through N, so KN: y = n + (45/n)x.

K is the intersection of KL and KN. Solving
−(n/45)x = n + (45/n)x
gives
x_K = −45 n^2/(n^2 + 45^2), y_K = n^3/(n^2 + 45^2).

The condition ∠KML = 50° means the slope of MK equals −tan(50°). Computing the slope of MK = (y_K − 0)/(x_K − 45) and equating gives the scalar equation
n^3 = 45·tan(50°)·(2 n^2 + 45^2).

Solving this numerically yields n ≈ 115.43. Then
K ≈ (x_K,y_K) = (−39.06, 100.20).

Finally compute the requested lengths:
- KN = distance between K and N = sqrt((x_K − 0)^2 + (y_K − n)^2) ≈ 41.92.
- KL = distance between K and L = sqrt(x_K^2 + y_K^2) ≈ 107.57.

Answers (rounded): KN ≈ 41.92, KL ≈ 107.57.