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Let me see.

Part (b)

a/sinA = c/sin C

asinC = csinA

asin17 = (2.2)(sin55)

a = (2.2)(sin55)/sin(17)

a = 6.2 miles

Part (c)

Dropping a perpendicular from angle C to D, I form a right triangle.

180° - 108° = 72°.

sin(72) = h/6.2

sin(72)(6.2) = h

5.89 miles = h

Part (d)

Let AC = b

b/sin(108) = 2.2/sin(17)

Solving for b, I get 7.2 miles from A to C.

I can now use the Pythagorean Theorem.

Let d = distance plane must travel before reaching A.

sin(35) = d/7.2

Solving for d I get 4.13 miles.

You say?

[QUOTE="nycmathguy, post: 1099884, member: 279"] Let me see. Part (b) a/sinA = c/sin C asinC = csinA asin17 = (2.2)(sin55) a = (2.2)(sin55)/sin(17) a = 6.2 miles Part (c) Dropping a perpendicular from angle C to D, I form a right triangle. 180° - 108° = 72°. sin(72) = h/6.2 sin(72)(6.2) = h 5.89 miles = h Part (d) Let AC = b b/sin(108) = 2.2/sin(17) Solving for b, I get 7.2 miles from A to C. I can now use the Pythagorean Theorem. Let d = distance plane must travel before reaching A. sin(35) = d/7.2 Solving for d I get 4.13 miles. You say? [/QUOTE]