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December 5th, 2016, 06:45 AM   #1
Joined: Apr 2016
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Really confused about this trigonometry question.

Ok so the question says...

In the triangle ABC, the sides AB and AC are equal in length. The side BC is 20 cm and angle BAC is 36 degrees. Find the area of the triangle.

I've tried this question numerous times and getting the wrong answer.

My method has been to split the triangle into half so I was working with a right-angled triangle with an angle of 18 degrees and a length of 10 cm at the bottom. However I feel I'm doing something wrong in the working out after that as I keep ending up with 3.25cm height? And using Base times height and dividing by 2, I would get the answer 32.5 which according to my book is wrong. The answer should apparently be 308cm...

I must be doing something wrong but I'm not sure what... Any help would be massively appreciated thanks!

Last edited by skipjack; December 5th, 2016 at 08:32 AM.
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December 5th, 2016, 08:16 AM   #2
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each congruent base angle in degrees has measure $\dfrac{180-36}{2} = 72$

height from the vertex angle to the base ...

$h = 10 \tan(72)$

$A = \dfrac{1}{2} b h = \dfrac{1}{2} \cdot 20 \cdot 10 \tan(72)$
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December 5th, 2016, 08:37 AM   #3
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Originally Posted by WarmCustard View Post
I keep ending up with 3.25cm height?
It seems you multiplied 10cm by tan(18°) instead of dividing 10cm by tan(18°).
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