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Determine which of the two acute angles has a cosine ofB 6 A8COS1011//CT/P

Determine Which Of The Two Acute Angles Has A Cosine OfB 6 A8COS1011CTP class=

Sagot :

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]

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