Question

What are the measures of all three anglesA ABC with side lengths 8, 11,and 17?

Answer 1

We know all the sides of the triangle. We need to find the missing angles.

To do this, we can use a formula known as the cosine rule:

`c=\sqrt[]{a^2+b^2-2ab\cos\gamma}`

Each one of the elements involved in this formula is explained in the next diagram:

Therefore, we can solve the equation above for gamma:

`\begin{gathered} c=\sqrt[]{a^2+b^2-2ab\cos\gamma} \\ \Rightarrow c^2-a^2-b^2=-2ab\cos \gamma \\ \Rightarrow a^2+b^2-c^2=2ab\cos \gamma \\ \Rightarrow\cos \gamma=(a^2+b^2-c^2)/(2ab) \end{gathered}`

Let's set a=8,b=11,c=17, then:

`\cos \gamma=-(104)/(176)=-(13)/(22)`

We can then use the arccos function to get the value of gamma:

`\begin{gathered} \cos \gamma=-(13)/(22) \\ \Rightarrow\gamma=\arccos (-(13)/(22))\approx126.2degree \end{gathered}`

The only option that has such an angle is the fourth one. Therefore, the answer is the fourth option: 126°, 32°, 22°