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SajjadAhmad
In ΔABC shown above, if ΔCAB equals 30°, then what is the area of the triangle?

A. 2
B. 2 \(\sqrt{2}\)
C. 2 \(\sqrt{3}\)
D. 3 \(\sqrt{2}\)
E. 4

Since all 3 angles in a triangle add to 180 degrees, we can conclude that ∠ACB is 60º
At this point, we should recognize that we have a SPECIAL 30-60-90 right triangle.
So, let's compare the given triangle with the BASE 30-60-90 right triangle




On the BASE triangle, we can see that the side opposite the 30º angle CORRESPONDS with side BC of the GIVEN triangle.
On the base triangle, this side has length 1, and on the given triangle, this side has length 2.
So, we can conclude that the given triangle is TWO TIMES the size of the base triangle.

So, the remaining two sides of the given triangle must be TWO TIMES the length of their CORRESPONDING sides on the base triangle:



What is the area of the given triangle?

Area = (base)(height)/2
= (2√3)(2)/2
= (4√3)/2
= 2√3

Answer: C
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Using trigonometry, the 30-60-90 triangle has sides in proportion to 1/2:sqrt(3)/2:1

(1/2)x4=2 Then we have to amplify by 4

Then the other side is sqrt(3)/2 x 4 = 2xsqrt(3)

And the area is 2x(2xsqrt(3))/2 = 2xsqrt(3)

So, C
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