__ you could our to descrimine a (i) determine to the nesres crivit a) b) __ 4.Brett needs to support a radio tower with guy wires. Each guy wire must run from the top of the tower to its own anchor 9 .00 m from the base of the tower. When the tower casts a shadow that is 9.00 m long.Brett's shadow is 060 m long. Brett is 1.85 m tall . What is the length of each guy wire that Brett needs? Comm 16 fun

## Step 1: Understand the problem and identify the triangle<br />### The problem involves a right triangle formed by the radio tower, its shadow, and the guy wire. The length of the guy wire corresponds to the hypotenuse of this triangle. We need to calculate the hypotenuse using the Pythagorean theorem.<br /><br />---<br /><br />## Step 2: Determine the height of the tower<br />### Brett's shadow and height form a similar triangle with the tower and its shadow. Using the property of similar triangles, we can set up the proportion:<br />\[<br />\frac{\text{Height of Brett}}{\text{Length of Brett's shadow}} = \frac{\text{Height of the tower}}{\text{Length of the tower's shadow}}<br />\]<br />Substitute the known values:<br />\[<br />\frac{1.85}{0.60} = \frac{\text{Height of the tower}}{9.00}<br />\]<br />Solve for the height of the tower:<br />\[<br />\text{Height of the tower} = \frac{1.85 \cdot 9.00}{0.60} = 27.75 \, \text{m}<br />\]<br /><br />---<br /><br />## Step 3: Apply the Pythagorean theorem<br />### The guy wire forms the hypotenuse of a right triangle where:<br />- One leg is the height of the tower ($27.75 \, \text{m}$),<br />- The other leg is the distance from the base to the anchor ($9.00 \, \text{m}$).<br /><br />Using the Pythagorean theorem:<br />\[<br />\text{Hypotenuse}^2 = (\text{Height of the tower})^2 + (\text{Distance to anchor})^2<br />\]<br />Substitute the values:<br />\[<br />\text{Hypotenuse}^2 = (27.75)^2 + (9.00)^2<br />\]<br />Calculate:<br />\[<br />\text{Hypotenuse}^2 = 770.0625 + 81 = 851.0625<br />\]<br />Take the square root to find the hypotenuse:<br />\[<br />\text{Hypotenuse} = \sqrt{851.0625} \approx 29.19 \, \text{m}<br />\]<br /><br />---