You can use a little trigonometry to find the sides of the right triangle, or you can simplify the resulting expression first. Let's do the latter, as it results in an interesting formula.
The area of a triangle is given by ...
... A = (1/2)bh
For this triangle, we can consider the base to be BC = AB·cos(∠B). The height is then AC = AB·sin(∠B) and the area is half the product of these:
... A = (1/2)·AB·cos(∠B)·AB·sin(∠B)
The product sin·cos is part of a double-angle trig identity, so we can write
.. A = (1/2)·(AB)²·(1/2)·sin(2·∠B)
... A = (1/4)·(AB)²·sin(2·∠B) . . . . . . area from hypotenuse and angle
Filling in the given numbers gives
... A = (1/4)·(12 cm)²·sin(2·75°)
... Area ∆ABC = 18 cm²
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Note that the above formula works using either acute angle in the right triangle. Using ∠B = 90°-∠A we can write it in terms of ∠A as
... A = (1/4)·(AB)²·sin(2·(90°-∠A)) = (1/4)·(AB)²·sin(180°-2·∠A)
Since sin(x) = sin(180°-x), this is the same as
... A = (1/4)·(AB)²·sin(2·∠A)
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Thank you for an interesting problem.
