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*Find the area of triangle ABC if AB = BC = 12 and angle ABC = 150 degrees.*

Since AB = BC triangle ABC is isoceles and angles BAC and BCA are equal

Since ABC = 150^{o} and the sum of the interior angles of a triangle is 180^{o}

then angles A and C total 30^{o} and therefore each of them is 15^{o}

Draw a perpendicular from the apex B down to a point on side AC - call the point P

BP

Using the sine function: sin(15) = ———

12

BP = (12)(0**.**65029) = 7**.**80 this is the height of the triangle

AP

Using the cosine function: cos(15) = ———

12

AP = (12)(0**.**96593) = 11**.**59 this is half the base of the triangle

So the whole base of the triangle is (2)(11**.**59) = 23**.**18 (We could have skipped this step. Do you know why?)

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The formula for the area of a triangle is — • base • height

_{ } 2

So, the area of this triangle is 0**.**5 • 23**.**18 • 7**.**80 = **90.40 units ^{2}**

The answer in the back of the book might be slightly different, because of the rounding that happened here.

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Guest Oct 30, 2020

edited by
Guest
Oct 30, 2020