Cosine Law Problems

A triangle has sides equal to 5 cm, 10 cm, and 7 cm. Find its angles (round answers to 1 decimal place).

Solution:

Set \( a = 10 \, \text{cm} \), \( b = 7 \, \text{cm} \), and \( c = 5 \, \text{cm} \).

• Use the cosine law to find the largest angle \( A \): \[ a^2 = b^2 + c^2 - 2bc \cos(A) \]
• Substitute \( a \), \( b \), and \( c \) by their values and solve for \( \cos(A) \): \[ \cos(A) = \frac{b^2 + c^2 - a^2}{2bc} \] \[ \cos(A) = \frac{7^2 + 5^2 - 10^2}{2 \cdot 7 \cdot 5} \] \[ \cos(A) = \frac{49 + 25 - 100}{70} = \frac{-13}{35} \]
• Use a calculator to find angle \( A \) and round to 1 decimal place: \[ A = \arccos\left(\frac{-13}{35}\right) \approx 111.8^\circ \]
• Use the sine law to find angle \( B \): \[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} \]
• Solve for \( \sin(B) \): \[ \sin(B) = \frac{b}{a} \sin(A) = \frac{7}{10} \sin(111.8^\circ) \]
• Use a calculator to find \( B \) and round to 1 decimal place: \[ B \approx 40.5^\circ \]
• Use the fact that the sum of all angles in a triangle is equal to \( 180^\circ \) to find angle \( C \): \[ C = 180^\circ - (40.5^\circ + 111.8^\circ) \approx 27.7^\circ \]

An aircraft tracking station determines the distance from a common point \( O \) to each aircraft and the angle between the aircrafts. If the angle \( O \) between the two aircrafts is \( 49^\circ \) and the distances from point \( O \) to the two aircrafts are 50 km and 72 km, find the distance \( d \) between the two aircrafts (round answers to 1 decimal place).

Solution:

• The cosine law can be used as follows: \[ d^2 = 72^2 + 50^2 - 2(72)(50) \cos(49^\circ) \]
• Solve for \( d \): \[ d = \sqrt{72^2 + 50^2 - 2(72)(50) \cos(49^\circ)} \approx 54.4 \, \text{km} \]

Question: A triangle has sides equal to 4 m, 11 m, and 8 m. Find its angles (round answers to 1 decimal place).

Solution:
\( A = 129.8^\circ \), \( B = 34.0^\circ \), \( C = 16.2^\circ \)

Question: A ship leaves port at 1 pm traveling north at the speed of 30 miles/hour. At 3 pm, the ship adjusts its course 20 degrees eastward. How far is the ship from the port at 4 pm? (round to the nearest unit).

Solution:
Distance traveled from 1 pm to 3 pm (2 hours at 30 mph) = 60 miles.
Distance traveled from 3 pm to 4 pm (1 hour at 30 mph) = 30 miles.
The angle between the northward path extension and the new course is \(20^\circ\), meaning the interior angle of the triangle is \(180^\circ - 20^\circ = 160^\circ\).
Using the cosine law:
\( d = \sqrt{60^2 + 30^2 - 2(60)(30)\cos(160^\circ)} \approx 89 \, \text{miles} \).