MHB Violet's Trigonometry Questions via Facebook

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The discussion focuses on solving various trigonometry problems using the sine and cosine rules. It demonstrates the application of the sine rule to find angles in triangles, specifically calculating the sine of angle β and using the cosine rule to determine angles in right-angled and non-right-angled triangles. The calculations include determining the lengths of sides and angles, such as finding the angle θ in an isosceles triangle and the area of a triangle using two sides and the included angle. The thread emphasizes the relationships between sides and angles in triangles, showcasing the importance of these rules in solving geometric problems. Overall, it provides clear examples of how to apply trigonometric principles effectively.
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1. By applying the sine rule we have

$\displaystyle \begin{align*} \frac{\sin{ \left( A \right) }}{a} &= \frac{\sin{ \left( B \right) } }{b} \\ \frac{\sin{ \left( \alpha \right) }}{10} &= \frac{\sin{ \left( \beta \right) }}{8} \\ \frac{\frac{2}{3}}{10} &= \frac{\sin{ \left( \beta \right) }}{8} \\ \frac{1}{15} &= \frac{\sin{ \left( \beta \right) }}{8} \\ \sin{ \left( \beta \right) } &= \frac{8}{15} \end{align*}$2. The length of the diagonal must be 10 units (it's a 3, 4, 5 right angled triangle magnified by a factor of 2 - check with Pythagoras if you'd like), so that means that the lengths from the centre to the vertices must each be 5 units in length.

Knowing this, the right hand isosceles triangle has lengths 5, 5, 6 and the angle between the two 5 lengths is $\displaystyle \begin{align*} \theta \end{align*}$, thus we can apply the cosine rule.

$\displaystyle \begin{align*} \cos{ \left( C \right) } &= \frac{a^2 + b^2 - c^2}{2\,a\,b} \\ \cos{ \left( \theta \right) } &= \frac{5^2 + 5^2 - 6^2}{2 \cdot 5 \cdot 5} \\ \cos{ \left( \theta \right) } &= \frac{25 + 25 - 36}{50} \\ \cos{ \left( \theta \right) } &= \frac{14}{50} \\ \cos{ \left( \theta \right) } &= \frac{7}{25} \\ \theta &= \cos^{-1}{\left( \frac{7}{25} \right) } \\ \theta &\approx 74^{\circ} \end{align*}$3. In $\displaystyle \begin{align*} \triangle ACD \end{align*}$ we can let $\displaystyle \begin{align*} \angle ACD = \theta \end{align*}$, then we can say that $\displaystyle \begin{align*} AD = \textrm{Opp} = 30 \end{align*}$ and $\displaystyle \begin{align*} CD = \textrm{Adj} = 40 \end{align*}$, so

$\displaystyle \begin{align*} \tan{ \left( \theta \right) } &= \frac{\textrm{Opp}}{\textrm{Adj}} \\ \tan{ \left( \theta \right) } &= \frac{30}{40} \\ \tan{ \left( \theta \right) } &= \frac{3}{4} \\ \theta &= \tan^{-1}{ \left( \frac{3}{4} \right) } \end{align*}$
 

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4. We have two sides and the angle between them, so we need to use the Cosine Rule.

$\displaystyle \begin{align*} b^2 &= a^2 + c^2 - 2\,a\,c\cos{ \left( B \right) } \\ b^2 &= 6^2 + 8^2 - 2 \cdot 6 \cdot 8 \cdot \cos{ \left( 120^{\circ} \right) } \\ b^2 &= 36 + 64 - 96\cos{ \left( 120^{\circ} \right) } \\ b^2 &= 100 - 96\cos{ \left( 120^{\circ} \right) } \\ b &= \sqrt{100 - 96\cos{ \left( 120^{\circ} \right) }} \end{align*}$5. The largest angle is opposite the largest side. You have all three sides of the triangle and are trying to find an angle, so you need to use the Cosine Rule.

$\displaystyle \begin{align*} \cos{ \left( C \right) } &= \frac{a^2 + b^2 - c^2}{2\,a\,b} \\ \cos{ \left( C \right) } &= \frac{6^2 + 5^2 - 10^2}{2 \cdot 6 \cdot 5} \\ \cos{ \left( C \right) } &= \frac{36 + 25 - 100}{60} \\ \cos{ \left( C \right) } &= \frac{-39}{\phantom{-}60} \\ \cos{ \left( C \right) } &= -\frac{13}{20} \\ C &= \cos^{-1}{ \left( -\frac{13}{20} \right) } \\ C &\approx 130^{\circ} \end{align*}$6. Again, we have all three sides and are trying to find an angle, so we need to use the Cosine Rule:

$\displaystyle \begin{align*} \cos{ \left( B \right) } &= \frac{a^2 + c^2 - b^2}{2\,a\,c} \\ \cos{ \left( B \right) } &= \frac{3^2 + 2^2 - 4^2}{2 \cdot 3 \cdot 2} \\ \cos{ \left( B \right) } &= \frac{9 + 4 - 16}{12} \\ \cos{ \left( B \right) } &= \frac{-3}{\phantom{-}12} \\ \cos{ \left( B \right) } &= -\frac{1}{4} \end{align*}$
 

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7. To find the area of a triangle, you can use two sides and the angle between them.

As the three angles in a triangle add to $\displaystyle \begin{align*} 180^{\circ} \end{align*}$, that means the unknown angle must be $\displaystyle \begin{align*} 180^{\circ} - 30^{\circ} - 50^{\circ} = 100^{\circ} \end{align*}$.

$\displaystyle \begin{align*} A &= \frac{1}{2} \,a\,b \sin{ \left( C \right) } \\ &= \frac{1}{2} \cdot 6.13\,\textrm{m} \cdot 4\,\textrm{m} \cdot \sin{ \left( 100^{\circ} \right) } \end{align*}$
 

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