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Violet's Trigonometry Questions via Facebook

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
week 3.jpg

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

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
week 3 part 2.JPG

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:

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
week 3 part 3.JPG

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 $\displaystyle \begin{align*} 180^{\circ} - 30^{\circ} - 50^{\circ} = 100^{\circ} \end{align*}$.

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