Question

two points on the opposite sides of a lake, d and e, are 355.5 and 276.2 ft, respectively, from a third point f on the shore. the lines joining points d and e with point f intersect at an angle of 71°45 (angle f-de). what is distance de?

Explanation:

Step1: Convert angle to decimal degrees

First, we convert \( 71^\circ 45' \) to decimal degrees. Since \( 1^\circ = 60' \), we have \( 45'=\frac{45}{60}=0.75^\circ \). So \( 71^\circ 45' = 71.75^\circ \).

Step2: Apply the Law of Cosines

The Law of Cosines states that for a triangle with sides \( a \), \( b \), and \( c \), and the angle \( C \) opposite side \( c \), \( c^{2}=a^{2}+b^{2}-2ab\cos(C) \). In triangle \( DEF \), let \( DF = a = 355.5 \) ft, \( EF = b = 276.2 \) ft, and the included angle \( \angle DFE = C=71.75^\circ \). We want to find the length of \( DE = c \).

Substitute the values into the Law of Cosines formula:
\[
c^{2}=(355.5)^{2}+(276.2)^{2}-2\times355.5\times276.2\times\cos(71.75^\circ)
\]

First, calculate \( (355.5)^{2}=355.5\times355.5 = 126380.25 \)
\( (276.2)^{2}=276.2\times276.2 = 76286.44 \)

Next, calculate \( 2\times355.5\times276.2=2\times355.5\times276.2 = 711\times276.2 = 196378.2 \)

Now, find \( \cos(71.75^\circ) \). Using a calculator, \( \cos(71.75^\circ)\approx0.3135 \)

Then, \( 2\times355.5\times276.2\times\cos(71.75^\circ)\approx196378.2\times0.3135\approx61564.56 \)

Now, substitute back into the formula for \( c^{2} \):
\[
c^{2}=126380.25 + 76286.44- 61564.56
\]
\[
c^{2}=126380.25+76286.44 = 202666.69
\]
\[
c^{2}=202666.69 - 61564.56=141102.13
\]

Step3: Find the square root of \( c^{2} \)

Take the square root of \( 141102.13 \) to find \( c \):
\[
c=\sqrt{141102.13}\approx375.6 \text{ ft}
\]

Answer:

The distance \( DE \) is approximately \( \boldsymbol{375.6} \) feet.