Elements of a triangle given its three sides

Text

A triangle has sides $a=9$, $b=8$, $c=13$.

(a) Find all the angles $\alpha,\beta,\gamma$ (opposite sides $a,b,c$ respectively).

(b) Find the altitude relative to side $a$.

(c) Find the median relative to side $b$.

(d) Find the bisector of angle $\gamma$.

Solution

(a) By the cosine rule: $\cos\alpha=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{152}{208}\approx0.731\Rightarrow\alpha\approx43.05°$; $\cos\beta=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{186}{234}\approx0.795\Rightarrow\beta\approx37.38°$; $\gamma=180°-\alpha-\beta\approx99.59°$ (check: $\cos\gamma=-\tfrac16$).

(b) By Heron’s formula: $s=15$, $\text{Area}=\sqrt{15\cdot6\cdot7\cdot2}=\sqrt{1260}\approx35.50$; so $h_a=\dfrac{2\,\text{Area}}{a}\approx7.89$.

(c) $m_b=\dfrac12\sqrt{2a^2+2c^2-b^2}=\dfrac12\sqrt{436}\approx10.44$.

(d) $t_\gamma=\dfrac{2ab\cos(\gamma/2)}{a+b}=\dfrac{144\cos49.80°}{17}\approx5.47$.

$\boxed{\alpha\approx43.05°,\ \beta\approx37.38°,\ \gamma\approx99.59°,\ h_a\approx7.89,\ m_b\approx10.44,\ t_\gamma\approx5.47}$