Question 956339
Find the area of ABC.

(A) = 50° ; AB = 15; and AC = 10   Area =

(B) = 75° ; CB = 12 ; and AB = 16   Area =

I know the Pythagorean Theorem is A^2 + B^2 = C^2, but I'm unsure how to use it in this case. 
What I did so far for A is add 15^2 (225) and 10^2 (100) to get 325. Am I doing it right?
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A^2 + B^2 = C^2 applies to right triangles only.
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(A) = 50° ; AB = 15; and AC = 10
Area = b*h/2
Using AB as the base, h = 15*sin(50)
Area = 10*15*sin(50)/2
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The 2nd problem is similar, just different numbers.
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PS  Use A for the angle and a for this side opposite.
CAPS for angles and lower case for sides.
{{{a^2 + b^2 = c^2}}} for right triangles.