Answer to Question #290910 in Geometry for twice

Question

Solve for α in the oblique triangle ABC; AB = 30; AC = 15 and angle B = 20°

Oblique triangle ABC; AB = 30; AC = 15 and angle B = 20°

Type out the two equations substituting the numbers from the diagram.

Type out the Law of Sines set of relationships and type out the most appropriate version to use the Law of Cosines for this solution.

Solve for a using both methods (show step by step work)

Answer 1

Using sine rule, we have;

$\displaystyle \frac{15}{\sin20^o}=\frac{30}{\sin\gamma}=\frac{a}{\sin\alpha}$

The first two equations yields;

$\displaystyle \sin\gamma=\frac{30\times\sin 20^o}{15}\\ \Rightarrow \gamma=\sin^{-1}\left(\frac{30\times\sin 20^o}{15}\right)=43.1601778^o$

but $\displaystyle \alpha +\gamma+20^o=180^o$

$\displaystyle \Rightarrow\alpha=180^o-20^o-\gamma=116.8398222^o$

Thus the above sine rule becomes;

$\displaystyle \frac{15}{\sin20^o}=\frac{30}{\sin 43.1601778^o}=\frac{a}{\sin116.8398222^o}$

The last two equations yields;

$\displaystyle a=\frac{30\times\sin116.8398222^o}{\sin43.1601778^o}=39.13244209$

Using Cosine rule;

$\displaystyle 15^2=30^2+a^2-2(30)(a)\cos20^o\\ \Rightarrow225=900+a^2-60a\cos20^o\\ \Rightarrow-675=a^2-60a\cos20^o\\ \Rightarrow a^2-60\cos20^oa+675\\ \Rightarrow a=39.13244209,\ 17.24911516$, (using quadratic formula).