Question

solve each triangle. round answers to the nearest tenth.
17)
18)
its a beautiful summer day and youre flying a kite. the angle of elevation between you and the kite is 57° and
the kite is 100 feet in the air. how long is your kite string?
from the top of a tower, the angle of depression from a handsome prince to a fair maiden is 72°. the top of the
tower is 80 feet above ground. how far from the castle is the maiden standing?

Explanation:

Step1: Find angle \( A \)

In triangle \( ABC \) (problem 17), since it is a right - triangle (\( \angle C = 90^{\circ} \)) and the sum of angles in a triangle is \( 180^{\circ} \), we have \( A=180^{\circ}-\angle B - \angle C \). Given \( \angle B = 62^{\circ} \) and \( \angle C = 90^{\circ} \), then \( A = 180^{\circ}-90^{\circ}-62^{\circ}=28^{\circ} \).

Step2: Find side \( BC \)

Using the trigonometric ratio \( \tan B=\frac{AC}{BC} \). We know \( AC = 22.6\space mi \) and \( B = 62^{\circ} \). So \( BC=\frac{AC}{\tan B}=\frac{22.6}{\tan62^{\circ}}\approx12.0\space mi \) (since \( \tan62^{\circ}\approx1.8807 \)).

Step3: Find side \( AB \)

Using the trigonometric ratio \( \sin B=\frac{AC}{AB} \). Then \( AB=\frac{AC}{\sin B}=\frac{22.6}{\sin62^{\circ}}\approx25.5\space mi \) (since \( \sin62^{\circ}\approx0.8829 \)).

For problem 18 (right - triangle \( ABC \) with \( \angle C = 90^{\circ} \), \( AC = 19\space in \), \( BC = 9\space in \)):

Step1: Find angle \( A \)

Using the trigonometric ratio \( \tan A=\frac{BC}{AC} \). So \( \tan A=\frac{9}{19}\approx0.4737 \), then \( A=\arctan(0.4737)\approx25.3^{\circ} \).

Step2: Find angle \( B \)

Since \( A + B+ C=180^{\circ} \) and \( C = 90^{\circ} \), \( B = 180^{\circ}-90^{\circ}-A \). So \( B=180^{\circ}-90^{\circ}-25.3^{\circ}=64.7^{\circ} \).

Step3: Find side \( AB \)

Using the Pythagorean theorem \( AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{19^{2}+9^{2}}=\sqrt{361 + 81}=\sqrt{442}\approx21.0\space in \).

For the kite problem:
Let the length of the kite string be \( l \). The height of the kite \( h = 100\space ft \) and the angle of elevation \( \theta=57^{\circ} \). Using the trigonometric ratio \( \sin\theta=\frac{h}{l} \). Then \( l=\frac{h}{\sin\theta}=\frac{100}{\sin57^{\circ}}\approx120.6\space ft \) (since \( \sin57^{\circ}\approx0.8387 \)).

For the tower problem:
Let the distance of the maiden from the castle be \( x \). The height of the tower \( h = 80\space ft \) and the angle of depression \( \theta = 72^{\circ} \). Since the angle of depression is equal to the angle of elevation from the maiden to the top of the tower. Using the trigonometric ratio \( \tan\theta=\frac{h}{x} \). So \( x=\frac{h}{\tan\theta}=\frac{80}{\tan72^{\circ}}\approx25.9\space ft \) (since \( \tan72^{\circ}\approx3.0777 \)).

Answer:

• Problem 17: \( A = 28^{\circ} \), \( BC\approx12.0\space mi \), \( AB\approx25.5\space mi \)
• Problem 18: \( A\approx25.3^{\circ} \), \( B\approx64.7^{\circ} \), \( AB\approx21.0\space in \)
• Kite problem: The kite string is approximately \( 120.6\space ft \) long.
• Tower problem: The maiden is approximately \( 25.9\space ft \) from the castle.