Hm... let me clarify, that's it?

I can't find any shortcut.

The length of the spring is given by l + e, where e = F/k, e = F/150

So, length of spring = (0.2 + F/150) m

Let Tc be the tension in the cord and Ts be the tension in the spring. (I'll use B as the angle instead)

Tc sinB + Ts sinC = 10

where C is the angle BCA.

Tc cosB = Ts cosC

From the cosine rule, we get:

AC^2 = BC^2 + AB^2 - 2(AB)(BC)cos B

(0.2 + F/150)^2 = 0.4^2 + 0.2^2 - 2(0.4)(0.2)cos B

(0.2 + F/150)^2 = 0.2 - 0.16cos B

Now, F = Ts

Using the first two equations, we get:

((Ts cosC)/cosB)sinB + Ts sinC = 10

Ts cosC tanB + Ts sinC = 10

Ts = 10/(cosC tanB + sinC)

Hence,

[0.2 + 10/{150(cosC tanB + sinC)}]^2 = 0.2 - 0.16cos B

Hm... not very 'pretty'. Let's try the sine rule.

[0.2 + 10/{150(cosC tanB + sinC)}/sinB = 0.2/sinC = 0.4/sinA

[0.2 + 10/{150(cosC tanB + sinC)} = (0.2 sinB)/sinC

So,

0.2 - 0.16cos B = ((0.2 sinB)/sinC)^2

0.2 - 0.16cos B = (0.04 sin^2B)/sin^2C

5 - 4cos B = sin^2B/sin^2C

We have:

5 - 4cos B = sin^2B/sin^2C

0.2/sinC = 0.4/sinA

A + B + C = 180

Hmm....

Maybe trial and error?