Given:
The right traingle ABC with side measurements.
Required:
[tex]\text{ Determine which of two acute angle has cosine of }\frac{4}{5}.[/tex]
Explanation:
First draw a figure
We need here cosine law
[tex]cosC=\frac{a^2+b^2-c^2}{2ab}[/tex]
So, we have a = 8, b = 10 and c = 6.
So, cosine of acute angle C
[tex]\begin{gathered} cosC=\frac{a^2+b^2-c^2}{2ab} \\ cosC=\frac{8^2+10^2-6^2}{2\times8\times10} \\ cosC=\frac{64+100-36}{160} \\ cosC=\frac{128}{160} \\ cosC=\frac{4}{5} \end{gathered}[/tex]
Answer:
[tex]\text{ Acute angle }C\text{ has cosine }\frac{4}{5}.[/tex]
