Circle with inscribed angle and arc

Question

I have a circle with an inscribed angle of 45°. The arc opposite the angle is marked (3w - 30)°. My instructions are Find w. Since the arc is always double the inscribed angle, I know that the inscribed angle is 90°. Obviously, 120 - 30 = 90, and 120 divided by 3 is 40°. Thus, w = 40. Every equation similar to this in my text book gives more information. Although I know the answer, I don't know how to reach it algebraically. I must show my work. What should I do?   

Answer 1

∠A is inscribed in arc BAC, intercepting arc BC

m∠A=(1/2)m(arc BC)

45º=(1/2)(3w-30)

multiply both sides by 2

2(45º)=2(1/2)(3w-30)

90º=(3w-30)

90=3w-30

90+30=3w-30+30

120=3w

120/3=3w/3

40=w

 

Another tutor named Russ P. did the problem on the 28th. Without a diagram it's difficult to determine which side of the triangle is part of the square base of the pyramid. Russ P. assumed it was 12 feet. The missing side of the triangle which is also the diameter of the semicircle is 12 feet; 13²=5²+x², 169-25=144 and √144=12 from the Pythagorean Theorem.

Volume=(1/3)*(area of the square base)*(height of the pyramid). Post again if you can provide more detail.