Question

name
challenge

1. application

in order to reach safely (probably), a roof constructor determined that she must place the base of her ladder 6 feet away from the house, making an angle of 76° with the ground. to the nearest tenth of a foot, how far above the ground is the eave of the roof?

1. challenge

find the measure of angle d

Explanation:

Problem 12 (Application)

Step 1: Identify the trigonometric relationship

We have a right triangle \( \triangle ABC \) with \( \angle A = 76^\circ \), \( AB = 6 \) ft (adjacent side to \( \angle A \)), and we need to find \( BC \) (opposite side to \( \angle A \)). We use the tangent function: \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)
So, \( \tan(76^\circ) = \frac{BC}{6} \)

Step 2: Solve for \( BC \)

Multiply both sides by 6: \( BC = 6 \times \tan(76^\circ) \)
Calculate \( \tan(76^\circ) \approx 4.0108 \)
Then \( BC \approx 6 \times 4.0108 \approx 24.06 \) ft

Step 1: Find the length of \( BC \)

In \( \triangle ABC \), \( \angle A = 30^\circ \), \( AC = 12 \) cm. We use the tangent function: \( \tan(30^\circ) = \frac{BC}{AC} \)
\( \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577 \)
So, \( BC = AC \times \tan(30^\circ) = 12 \times \frac{1}{\sqrt{3}} = 4\sqrt{3} \) cm (or approximately \( 6.928 \) cm)

Step 2: Find \( \angle D \) in \( \triangle BCD \)

In \( \triangle BCD \), \( BC = 4\sqrt{3} \) cm, \( CD = 4 \) cm. We use the tangent function: \( \tan(\angle D) = \frac{BC}{CD} \)
Substitute \( BC = 4\sqrt{3} \) and \( CD = 4 \): \( \tan(\angle D) = \frac{4\sqrt{3}}{4} = \sqrt{3} \)
We know that \( \tan(60^\circ) = \sqrt{3} \), so \( \angle D = 60^\circ \)

Answer:

Approximately \( 24.06 \) feet above the ground.

Problem 13 (Challenge)