Question

for a triangle with the sides of lengths of 7, 9 and 14, find to 1 decimal place the measure of: i) the smallest angle ii) the largest angle

Explanation:

Step1: Recall the cosine - law

The cosine - law is $c^{2}=a^{2}+b^{2}-2ab\cos C$, where $a$, $b$, $c$ are the sides of the triangle and $C$ is the angle opposite to side $c$. The smallest angle is opposite the smallest side and the largest angle is opposite the largest side. Let $a = 7$, $b = 9$, $c = 14$.

Step2: Find the smallest angle (opposite side $a$)

Using the cosine - law $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}$. Substitute $a = 7$, $b = 9$, $c = 14$:
\[

$$\begin{align*} \cos A&=\frac{9^{2}+14^{2}-7^{2}}{2\times9\times14}\\ &=\frac{81 + 196-49}{252}\\ &=\frac{228}{252}\approx0.90476 \end{align*}$$

\]
Then $A=\cos^{-1}(0.90476)\approx25.2^{\circ}$

Step3: Find the largest angle (opposite side $c$)

Using the cosine - law $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$. Substitute $a = 7$, $b = 9$, $c = 14$:
\[

$$\begin{align*} \cos C&=\frac{7^{2}+9^{2}-14^{2}}{2\times7\times9}\\ &=\frac{49 + 81-196}{126}\\ &=\frac{130 - 196}{126}\\ &=-\frac{66}{126}\approx - 0.52381 \end{align*}$$

\]
Then $C=\cos^{-1}(-0.52381)\approx121.6^{\circ}$

Answer:

i) $25.2^{\circ}$
ii) $121.6^{\circ}$