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Thread: iscoeles triangle

  1. #1
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    Exclamation iscoeles triangle

    I know that a isoceles triangle has 2 equal sides and 2 equal angles. The sides of the 2 sides are 5m and the angles are 72.64 degrees, making the top angle 34.72 degrees. But i do not know how to work out the base length? can you help me please
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  2. #2
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    Quote Originally Posted by kath View Post
    I know that a isoceles triangle has 2 equal sides and 2 equal angles. The sides of the 2 sides are 5m and the angles are 72.64 degrees, making the top angle 34.72 degrees. But i do not know how to work out the base length? can you help me please
    use the sine rule:

    $\displaystyle \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{sin(C)}$

    where side a is opposite angle A, side b opposite angle B and side c opposite angle C.

    I'm going to define the side you need to find as side a and so A = 34.72

    $\displaystyle \frac{a}{\sin(A)} = \frac{b}{\sin(B)}$ thus $\displaystyle \frac{a}{sin(34.72)} = \frac{5}{sin(72.64)}$

    this gives

    $\displaystyle a = \frac{5sin(34.72)}{sin(72.64)} = 2.98m (3sf)$
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  3. #3
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    law of cosines ... $\displaystyle a^2 = b^2 + c^2 - 2 \cdot b \cdot c \cdot \cos{A}$

    let $\displaystyle y$ = base length

    $\displaystyle x$ = length of the two equal sides

    $\displaystyle Y$ = measure of the vertex angle

    $\displaystyle y^2 = x^2 + x^2 - 2 \cdot x \cdot x \cdot \cos{Y}$

    $\displaystyle y^2 = 2x^2(1 - \cos{Y})$

    $\displaystyle y = \sqrt{2x^2(1 - \cos{Y})}$
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