# Chapter 1 - Trigonometric Functions - Section 1.2 Angle Relationships and Similar Triangles - 1.2 Exercises - Page 19: 67

$c=111\frac{1}{9}$

#### Work Step by Step

1. The ratio is the long leg over the hypotenuse for each of the two triangles. Don't forget that you need to do the same addition to get the long leg of the larger triangle. $\frac{90}{100}=\frac{10+90}{c}$ 2. Solve for c: $9c=1000$ $c=111\frac{1}{9}$

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