Question

1. boat navigation (law of sines)

two boats leave a dock and travel in different
directions. one boat travels 40 km, the other
55 km, and the angle between their paths is
63°.
a) find the distance between the two boats
using the law of cosines.
b) if the angle opposite the 55 km path is
measured, find it using the law of sines.

Explanation:

Part (a)

Step1: Recall Law of Cosines

The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and the angle \(C\) opposite side \(c\), \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). Let the two sides of the triangle (distances traveled by boats) be \(a = 40\) km, \(b=55\) km, and the included angle \(C = 63^{\circ}\). We need to find the third side \(c\) (distance between the boats).

Step2: Substitute values into the formula

Substitute \(a = 40\), \(b = 55\), and \(C=63^{\circ}\) into the Law of Cosines formula:

$$
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First, calculate \(\cos(63^{\circ})\approx0.4540\)

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

Take the square root of \(2627.4\) to find \(c\):
\(c=\sqrt{2627.4}\approx51.26\) km

Part (b)

Step1: Recall Law of Sines

The Law of Sines states that \(\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}\). Let \(a = 55\) km, \(c\approx51.26\) km (from part a), and \(C = 63^{\circ}\). We need to find the angle \(A\) (opposite the \(55\) km path).

Step2: Substitute values into the formula

From \(\frac{a}{\sin(A)}=\frac{c}{\sin(C)}\), we can solve for \(\sin(A)\):
\(\sin(A)=\frac{a\times\sin(C)}{c}\)
Substitute \(a = 55\), \(c\approx51.26\), and \(C = 63^{\circ}\) (\(\sin(63^{\circ})\approx0.8910\)):

$$
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Step3: Find the angle \(A\)

Take the inverse sine of \(0.9558\) to find \(A\):
\(A=\sin^{-1}(0.9558)\approx73^{\circ}\) (we can also check for the other possible angle \(180 - 73=107^{\circ}\), but since \(a = 55>c\approx51.26\), angle \(A\) should be greater than \(C = 63^{\circ}\), and both \(73^{\circ}\) and \(107^{\circ}\) are possible, but in the context of boat navigation (acute angle between paths), \(73^{\circ}\) is more likely. However, if we consider the triangle, the sum of angles in a triangle is \(180^{\circ}\), and if we assume the triangle is valid, we can calculate the angle. But from the calculation \(\sin(A)\approx0.9558\), \(A\approx73^{\circ}\) or \(A\approx107^{\circ}\). But since \(a = 55\) and \(c\approx51.26\), and \(A\) is opposite \(a\), and \(C = 63^{\circ}\), if \(A = 107^{\circ}\), then the third angle \(B=180-(107 + 63)=10^{\circ}\), which is also valid. But usually, in such problems, we consider the acute angle or the angle that makes sense in the context. But from the calculation, \(\sin(A)\approx0.9558\), so \(A\approx73^{\circ}\) (using a calculator, \(\sin^{-1}(0.9558)\approx73^{\circ}\))

Final Answers

a) The distance between the two boats is approximately \(\boxed{51.3}\) km (rounded to one decimal place)
b) The angle opposite the \(55\) km path is approximately \(\boxed{73^{\circ}}\) (or \(107^{\circ}\), but \(73^{\circ}\) is more likely in the context)

Answer:

Step1: Recall Law of Sines

The Law of Sines states that \(\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}\). Let \(a = 55\) km, \(c\approx51.26\) km (from part a), and \(C = 63^{\circ}\). We need to find the angle \(A\) (opposite the \(55\) km path).

Step2: Substitute values into the formula

From \(\frac{a}{\sin(A)}=\frac{c}{\sin(C)}\), we can solve for \(\sin(A)\):
\(\sin(A)=\frac{a\times\sin(C)}{c}\)
Substitute \(a = 55\), \(c\approx51.26\), and \(C = 63^{\circ}\) (\(\sin(63^{\circ})\approx0.8910\)):

$$
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$$

Step3: Find the angle \(A\)

Take the inverse sine of \(0.9558\) to find \(A\):
\(A=\sin^{-1}(0.9558)\approx73^{\circ}\) (we can also check for the other possible angle \(180 - 73=107^{\circ}\), but since \(a = 55>c\approx51.26\), angle \(A\) should be greater than \(C = 63^{\circ}\), and both \(73^{\circ}\) and \(107^{\circ}\) are possible, but in the context of boat navigation (acute angle between paths), \(73^{\circ}\) is more likely. However, if we consider the triangle, the sum of angles in a triangle is \(180^{\circ}\), and if we assume the triangle is valid, we can calculate the angle. But from the calculation \(\sin(A)\approx0.9558\), \(A\approx73^{\circ}\) or \(A\approx107^{\circ}\). But since \(a = 55\) and \(c\approx51.26\), and \(A\) is opposite \(a\), and \(C = 63^{\circ}\), if \(A = 107^{\circ}\), then the third angle \(B=180-(107 + 63)=10^{\circ}\), which is also valid. But usually, in such problems, we consider the acute angle or the angle that makes sense in the context. But from the calculation, \(\sin(A)\approx0.9558\), so \(A\approx73^{\circ}\) (using a calculator, \(\sin^{-1}(0.9558)\approx73^{\circ}\))

Final Answers

a) The distance between the two boats is approximately \(\boxed{51.3}\) km (rounded to one decimal place)
b) The angle opposite the \(55\) km path is approximately \(\boxed{73^{\circ}}\) (or \(107^{\circ}\), but \(73^{\circ}\) is more likely in the context)