Solve for a and b in the given triangle.

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Solution:

Since the sum of the angles of the triangle is 180^o, we get that B = 60^o. Now we can find a by using the trigonometric ratio sin = opp/hyp. Therefore:

sin30^o = a/12
a = 12 * sin30^o = 12(1/2) = 6

To find b, we can use cos30^o or sin60^o, we get the same answer. Let's use

cos30^o = b/12
b = 12 * cos30^o = 12(/2) = 6

Example 2:

Find the value of sin240^o and cos495^o.

Solution:

The reference angle is going to be 240^o - 180^o = 60^o. But with a 240^o angle, sin is negative. Therefore we get:

sin240^o = -sin60^o = -/2

For 495^o the coterminal angle is 135^o. The reference angle is 180^o - 135^o = 45^o. Since the sign of cos135^o is negative, we get:

cos495^o = cos135^o = -cos45^o = -/2

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