Question

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to find the distance across a small lake, a surveyor has taken the measurements shown. find the distance across the lake using this information.

note: the triangle is not drawn to scale.
distance =
enter your answer as a number; your answer should be accurate to 2 decimal places.
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Explanation:

Step1: Identify the Law to Use

We have a triangle \( \triangle ABC \) with sides \( AC = 2.07 \) km, \( BC = 1.18 \) km, and angle \( C = 46.4^\circ \). We need to find the length of \( AB \) (the distance across the lake). We can use the Law of Cosines, which states that for a triangle with sides \( a \), \( b \), \( c \) and the angle \( C \) opposite side \( c \), \( c^{2}=a^{2}+b^{2}-2ab\cos(C) \). Here, \( AB \) is opposite angle \( C \), \( AC = b = 2.07 \), \( BC=a = 1.18 \), and \( C = 46.4^\circ \).

Step2: Apply the Law of Cosines

Let \( AB = c \). Then:

$$ c^{2}=1.18^{2}+2.07^{2}-2\times1.18\times2.07\times\cos(46.4^\circ) $$

First, calculate the squares and the product:
\( 1.18^{2}=1.3924 \), \( 2.07^{2} = 4.2849 \), and \( 2\times1.18\times2.07=2\times2.4426 = 4.8852 \)
Now, find \( \cos(46.4^\circ)\approx\cos(46.4)\approx0.6909 \)
Then:

$$ c^{2}=1.3924 + 4.2849-4.8852\times0.6909 $$

Calculate \( 4.8852\times0.6909\approx4.8852\times0.6909\approx3.375 \)
Then \( c^{2}=1.3924 + 4.2849- 3.375=5.6773 - 3.375 = 2.3023 \)
Take the square root: \( c=\sqrt{2.3023}\approx1.52 \)

Answer:

\( 1.52 \)