The problem consists of three sub-problems: 1. Find the measure of the complementary angle of an angle measuring $69.1^\circ$.

1. Problem Description

The problem consists of three sub-problems:

1. Find the measure of the complementary angle of an angle measuring $69.1^\circ$.

2. Find the length of the hypotenuse of a right triangle with legs of length 45 miles and 60 miles. Round the answer to the nearest tenth.

3. Determine if a triangle can have sides of lengths 4, 12, and

1

8.

2. Solution Steps

1. Complementary angles are two angles whose measures add up to $90^\circ$. Let $x$ be the measure of the complementary angle. Then $x + 69.1^\circ = 90^\circ$. So, $x = 90^\circ - 69.1^\circ = 20.9^\circ$.

2. We are given a right triangle with legs of length 45 miles and 60 miles. Let $c$ be the length of the hypotenuse. By the Pythagorean theorem, we have $c^2 = a^2 + b^2$, where $a$ and $b$ are the lengths of the legs. Thus, $c^2 = 45^2 + 60^2 = 2025 + 3600 = 5625$. Taking the square root of both sides gives $c = \sqrt{5625} = 75$.

3. For a triangle with sides of lengths $a$, $b$, and $c$, the triangle inequality must hold: $a + b > c$, $a + c > b$, and $b + c > a$.

In our case, $a = 4$, $b = 12$, and $c = 18$. We have:

$4 + 12 = 16$, which is not greater than

1

8. Thus, $4 + 12 \not{>} 18$.

$4 + 18 = 22 > 12$.

$12 + 18 = 30 > 4$.

Since the triangle inequality does not hold for all three conditions, a triangle with side lengths 4, 12, and 18 cannot exist.

3. Final Answer

1. $20.9^\circ$

2. $75$ miles

3. no