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jmc

geometry junior

Problem

In right triangle

A B C

, shown below,

cos C = \frac{9 130}{130}

. Find

A C

.

problem

Solution

Since

cos C = \frac{9 130}{130}

and

cos C = \frac{A C}{B C}

, we have

\frac{A C}{B C} = \frac{A C}{130} = \frac{9 130}{130}

. This means that

A C = \frac{9 130}{130} \cdot 130 = \frac{9 \cdot 130 \cdot 130}{130} = 9

.

Final answer

9