Items in Green and Violet were added to aid in the solution and do not appear on the original diagram.

First, identify the essential nodes, label one as the reference node and label all remaining essential nodes with node-voltage variables. This is shown in Green on the diagram above.

is = V_A / 50

Referring to the diagram above, it is easily seen that KCL at node A yields

I5 + is + I10 = 0

The main difficulty remaining is to define these currents in terms of the node voltages.

First, is is simple and was already determined above.

For I5 we need to determine V5 first. By KVL, we find that

VA = V5 + 80

so

V5 = VA - 80

thus

I5 = V5 / 5 = (VA - 80) / 5

Finally, the hardest one. Referring to the redrawn circuit diagram below, note that if we replace the components inside the Blue dotted outline with something completely equivalent, the circuit outside the dotted outline will behave identically. Since I10 is outside the dotted outline, it will be the same if we replace the series combination of resistor - dependent source - resistor with a simpler equivalent circuit.

Since the three components inside the dotted blue box are all in series, the two resistors (10 W and 15 W) are actually in series although the dependent source is between them (they have the same current through them). Thus we can combine these two resistors in series giving the following slightly simpler circuit:

Now it is considerably easier to determine the voltage across the equivalent 25 W resistor, thus I10.
By KVL

VA = V₂5 - 75 is

Inserting the expression for is from above gives

VA = V₂5 - 75 (V_A / 50) = V₂5 - 1.5 V_A

thus

V25 = V_A + 1.5 V_A = 2.5 V_A

and so

I10 = V₂5 / 25 = 2.5 VA / 25 = 0.1 V_A

Inserting all these currents into the KCL equation, we get

(V_A - 80) / 5 + V_A / 50 + 0.1 V_A = 0
0.2 V_A - 16 + 0.02 V_A + 0.1 V_A = 0
0.32 V_A = 16
V_A = 50

Now that we have the node voltage , we can calculate all other parameters in the circuit relatively easily. Using the expressions determined above, we find that

is = V_A / 50 = 50 /50 = 1 A
I10 = 0.1 V_A = 0.1 (50) = 5 A

and the voltage across the dependent source is thus

75 is = 75 V

The problem asked for power delivered (or generated) by the dependent source. We know that for power delivered is

P = VI (no negative sign)

if the arrow enters the negative (one of the few exceptions to the arrow entering the positive rule) as it does on the diagram. Therefore, the power delivered by the dependent source is

P = (75 is) (I10) = (75 V) (5 A)
P = 375 W