Question

1. find each answer to the nearest tenth.

a. two lighthouses are located 12.0 miles apart. a ship is observed between the two at point f, and an observer at lighthouse d notes that ∠fde measures 70°. an observer in lighthouse e notes that ∠fed measures 23°. how far is the ship from each lighthouse?
b. an isosceles triangle has base of length 24 meters. if the vertex angle measures 54°, what is the perimeter of the triangle?

Explanation:

Part a

Step1: Find the third angle in triangle \( DEF \)

In triangle \( DEF \), the sum of interior angles is \( 180^\circ \). Given \( \angle FDE = 70^\circ \) and \( \angle FED = 23^\circ \), we find \( \angle DFE \) as:
\( \angle DFE=180^\circ - 70^\circ - 23^\circ = 87^\circ \)

Step2: Apply the Law of Sines

The Law of Sines states that \( \frac{DF}{\sin\angle FED}=\frac{EF}{\sin\angle FDE}=\frac{DE}{\sin\angle DFE} \), where \( DE = 12.0 \) miles (distance between lighthouses).

First, find \( DF \):
\( \frac{DF}{\sin23^\circ}=\frac{12.0}{\sin87^\circ} \)
\( DF=\frac{12.0\times\sin23^\circ}{\sin87^\circ} \)
\( \sin23^\circ\approx0.3907 \), \( \sin87^\circ\approx0.9986 \)
\( DF\approx\frac{12.0\times0.3907}{0.9986}\approx4.7 \) miles

Next, find \( EF \):
\( \frac{EF}{\sin70^\circ}=\frac{12.0}{\sin87^\circ} \)
\( EF=\frac{12.0\times\sin70^\circ}{\sin87^\circ} \)
\( \sin70^\circ\approx0.9397 \)
\( EF\approx\frac{12.0\times0.9397}{0.9986}\approx11.3 \) miles

Part b

Step1: Analyze the isosceles triangle

Let the isosceles triangle be \( ABC \) with base \( BC = 24 \) meters and vertex angle \( \angle BAC = 54^\circ \). The base angles \( \angle ABC \) and \( \angle ACB \) are equal. Each base angle is \( \frac{180^\circ - 54^\circ}{2}=63^\circ \)

Step2: Apply the Law of Sines

Let the equal sides be \( AB = AC = x \). By the Law of Sines: \( \frac{BC}{\sin\angle BAC}=\frac{AB}{\sin\angle ACB} \)
\( \frac{24}{\sin54^\circ}=\frac{x}{\sin63^\circ} \)
\( x = \frac{24\times\sin63^\circ}{\sin54^\circ} \)
\( \sin54^\circ\approx0.8090 \), \( \sin63^\circ\approx0.8910 \)
\( x=\frac{24\times0.8910}{0.8090}\approx26.3 \) meters

Step3: Calculate the perimeter

Perimeter \( P=AB + AC + BC=2x + 24 \)
\( P = 2\times26.3+24=52.6 + 24 = 76.6 \) meters

Part a Answer:

Distance from ship to lighthouse \( E \) ( \( DF \) ) is approximately \( \boldsymbol{4.7} \) miles, and to lighthouse \( D \) ( \( EF \) ) is approximately \( \boldsymbol{11.3} \) miles.

Part b Answer:

The perimeter of the isosceles triangle is approximately \( \boldsymbol{76.6} \) meters.

Answer:

Step1: Analyze the isosceles triangle

Let the isosceles triangle be \( ABC \) with base \( BC = 24 \) meters and vertex angle \( \angle BAC = 54^\circ \). The base angles \( \angle ABC \) and \( \angle ACB \) are equal. Each base angle is \( \frac{180^\circ - 54^\circ}{2}=63^\circ \)

Step2: Apply the Law of Sines

Let the equal sides be \( AB = AC = x \). By the Law of Sines: \( \frac{BC}{\sin\angle BAC}=\frac{AB}{\sin\angle ACB} \)
\( \frac{24}{\sin54^\circ}=\frac{x}{\sin63^\circ} \)
\( x = \frac{24\times\sin63^\circ}{\sin54^\circ} \)
\( \sin54^\circ\approx0.8090 \), \( \sin63^\circ\approx0.8910 \)
\( x=\frac{24\times0.8910}{0.8090}\approx26.3 \) meters

Step3: Calculate the perimeter

Perimeter \( P=AB + AC + BC=2x + 24 \)
\( P = 2\times26.3+24=52.6 + 24 = 76.6 \) meters

Part a Answer:

Distance from ship to lighthouse \( E \) ( \( DF \) ) is approximately \( \boldsymbol{4.7} \) miles, and to lighthouse \( D \) ( \( EF \) ) is approximately \( \boldsymbol{11.3} \) miles.

Part b Answer:

The perimeter of the isosceles triangle is approximately \( \boldsymbol{76.6} \) meters.