Question 956339: I'm having some trouble with Algebra 2. The problem is:
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?
Thanks for the help, I appreciate it!
Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 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.
 for right triangles.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! Find the area of ABC.
(A) = 50° ; AB = 15; and AC = 10 Area =
(B) = 75° ; CB = 12 ; and AB = 16 Area =
The area of a triangle is equal to 1/2 * the base * the height.
if you let b = the base and h = the height, and a equal the area, then the formula becomes:
a = 1/2 * b * h
you are not given the height so you have to find it.
you also need to re-orient your triangles so that the given angle is not opposite the base of the triangle.
this will allow you to solve for h easily.
in the first triangle, make A bottom left and B top middle and C bottom right and then draw h from B intersecting with AC at D forming a 90 degree angle with AC. This makes h the height of the triangle and it makes AC the base of the triangle.
in the second triangle, make C bottom left and A middle top and B bottom right and then draw h from A intersecting with CB at D forming a 90 degree angle with CB. This makes h the height of the triangle and it makes CB the base of the triangle.
Now you are ready to solve.
First you need to find h.
In the first triangle, sin(50) = opp/hyp = h/15.
solve for h to get h = 15 * sin(50).
In the second triangle, sin(75) = opp/hyp = h/16.
solve for h to get h = 16 * sin(75).
the formula for the area of a triangle is a = 1/2 * b * h
a is the area.
b is the base.
h is the height.
in the first triangle, you get:
a = 1/2 * b * h becomes:
a = 1/2 * 10 * 15 * sin(50).
in the second triangle, you get:
a = 1/2 * b * h becomes:
a = 1/2 * 12 * 16 * sin(75).
use your calculator to solve:
in the first triangle you get a = 57.45 square units.
in the second triangle you get a = 92.73 square units.
results are rounded to 2 decimal places.
a picture of your triangles is shown below:
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