Using the Law of Sines and Cosines to Find Angles

Using the Law of Sines and Cosines to Find Angles exercise

• Calculate the angle using the law of sines.

  Hints

  You can use opposite operations to solve this equation.

  Start by substituting in the given values.

  Not all given values will be needed.

  Solution

  Start with the given equation from the law of sines:

  $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$.

  Substitute in the given values:

  $\dfrac{5.18}{\sin 33.4°}=\dfrac{9}{\sin B}$.

  Use opposite operations to begin to isolate B:

  $5.18(\sin B)=9(\sin 33.4°)$

  $\sin B=\dfrac{9(\sin 33.4°)}{5.18}$.

  Apply the inverse $\sin$ to both sides,

  $B=\sin^{-1}(\dfrac{9(\sin 33.4°)}{5.18})$,

  to get $B=73°$.

• Find angle $A$ using the law of cosines.

  Hints

  The angle $A$ is written with a capital $A$, while the side $a$ is written with a lowercase $a$.

  Before isolating $A$ completely, Pete needs to isolate $\cos A$.

  To convert $\cos A$ into $A$, Pete should apply the inverse cosine operation.

  Solution

  The law of cosines states that:

  $a^2 = b^2 + c^2 - 2(bc)\cos A$

  $b^2 = a^2 + c^2 - 2(ac)\cos B$

  $c^2 = a^2 + b^2 - 2(ab)\cos C$

  The first equation is the only one that includes the angle $A$, so this is the equation Pete must use.

  Pete must use opposite operations to isolate $\cos A$:

  $b^2 + c^2 - a^2 = 2(bc)\cos A$

  $\dfrac{b^2 + c^2 - a^2}{2(bc)} = \cos A$

  Next he must apply the inverse cosine operation to both sides. This will completely isolate $A$ on the right hand side.

  $\cos^{-1}(\dfrac{b^2 + c^2 - a^2}{2(bc)}) = A$

  Substituting in known values, Pete finds:

  $\cos^{-1}(\dfrac{9^2 + 6^2 - 5.18^2}{2(9)(6)}) = A$

  $A = 33.4°$

• Determine the law of cosines equation.

  Hints

  It will help to begin by writing down the law of cosines equation that includes angle $A$, and then to rearrange this equation to isolate $A$.

  It may help you to assign the labels $a$, $b$, and $c$ to the side lengths of each triangle.

  Solution

  We must find an equation that shows the angle $A$ in terms of the side lengths. We can do this using the law of cosines, which states that:

  $a^2 = b^2 + c^2 - 2(bc)\cos A$

  $b^2 = a^2 + c^2 - 2(ac)\cos B$

  $c^2 = a^2 + b^2 - 2(ab)\cos C$

  In each case we are looking for angle $A$. The first equation is the only one that includes the angle $A$, so this is the equation that we must use.

  We must first isolate $\cos A$. We can do this using opposite operations:

  $b^2 + c^2 - a^2 = 2(bc)\cos A$

  $\dfrac{b^2 + c^2 - a^2}{2(bc)} = \cos A$

  Next he must apply the inverse cosine operation to both sides. This will completely isolate $A$ on the right hand side.

  $\cos^{-1}(\dfrac{b^2 + c^2 - a^2}{2(bc)}) = A$.

  Now we can substitute our known values for each triangle into this equation.

  Recall that the side length $a$, is the side opposite angle $A$. Therefore for the first triangle:

  $a = 5.18$

  $b = 9$

  $c = 6$

  Substituting these values into the equation we found above yields:

  $cos^{-1}(\dfrac{9^2 + 6^2 - 5.18^2}{2(9)(6)}) = A$.

  Similarly, for the second triangle:

  $a = 7$

  $b = 9$

  $c = 8$

  $cos^{-1}(\dfrac{9^2 + 8^2 - 7^2}{2(9)(8)}) = A$.

  For the third triangle:

  $a = 2$

  $b = 3$

  $c = 4$

  $cos^{-1}(\dfrac{3^2 + 4^2 - 2^2}{2(3)(4)}) = A$.

  For the fourth triangle:

  $a = 7$

  $b = 8$

  $c = 9$

  $cos^{-1}(\dfrac{8^2 + 9^2 - 7^2}{2(8)(9)}) = A$.

• Solve for the angle Pete needs to navigate to the island.

  Hints

  Think about how you can use the law of cosines to find the missing angle.

  Since the points on this triangle are $A$, $C$, and $D$, it may help to substitute these variable names into the law of cosines equation.

  Remember that the side length is named by the angle it is opposite to. For example, side $a$ is opposite angle $A$.

  Solution

  Pete must find angle $A$, given all side lengths of a triangle.

  The law of cosines states that:

  $a^2 = b^2 + c^2 - 2(bc)\cos A$

  $b^2 = a^2 + c^2 - 2(ac)\cos B$

  $c^2 = a^2 + b^2 - 2(ab)\cos C$

  The first equation is the only one that includes the angle $A$, so this is the equation that Pete must use.

  Pete's triangle has points $A$, $C$, and $D$, so he decides to substitute $d$ in for $b$ in the equation:

  $a^2 = d^2 + c^2 - 2(dc)\cos A$

  Pete must use opposite operations to isolate $\cos A$:

  $d^2 + c^2 - a^2 = 2(dc)\cos A$

  $\dfrac{d^2 + c^2 - a^2}{2(dc)} = \cos A$

  Next he must apply the inverse cosine operation to both sides. This will completely isolate $A$ on the right hand side.

  $cos^{-1}(\dfrac{d^2 + c^2 - a^2}{2(dc)}) = A$

  Pete knows that the side a is opposite angle $A$ in a triangle. He determines the side lengths of this triangle to be:

  $a = 4$

  $d = 9$

  $c = 6.5$

  Substituting in known values, Pete finds that

  $cos^{-1}(\dfrac{9^2 + 6.5^2 - 4^2}{2(9)(6.5)}) = A$.

  So $A = 30.3°$.

• Define the law of cosines.

  Hints

  Pay close attention to the signs.

  Only one of the equations is correct.

  It may help to write out the equations of the law of cosines, and then compare these equations to the ones given.

  Solution

  The equations given by the law of cosines are:

  1. $a^2 = b^2 + c^2 - 2(bc)\cos A$
  2. $b^2 = a^2 + c^2 - 2(ac)\cos B$
  3. $c^2 = a^2 + b^2 - 2(ab)\cos C$

  Let's compare the given equations to these:

  ✓ $a^2 = b^2 + c^2 - 2(bc)\cos A$

  This equation matches equation $1$ exactly.

  ✗ $b^2 = a^2 - c^2 - 2(ac)\cos B$

  This equation almost matches equation $2$, but the sign on the $c^2$ term is wrong.

  ✗ $c^2 = a^2 + b^2 + 2(ab)\cos C$

  This equation almost matches equation $3$, but the sign on the $2(ab)\cos C$ term is wrong.

  ✗ $a^2 = b^2 + c^2 - 2(ac)\cos A$

  This equation almost matches equation $1$, but the term multiplied by $\cos a$ should be $-2(bc)$ instead of $-2(ac)$.

• Identify the values of the angles marking potential new locations for lighthouses.

  Hints

  Look at the equations given by the law of sines and law of cosines. Which equations contain the unknown value, and only other known values?

  Be careful with signs as you use opposite operations to isolate the required variable.

  Solution

  Pete determines that to find angle $B$, he should use the law of cosines, which includes one angle and all side lengths of a triangle. Pete knows that the three equations given by the law of cosines are:

  1. $a^2 = b^2 + c^2 - 2(bc)\cos A$
  2. $b^2 = a^2 + c^2 - 2(ac)\cos B$
  3. $c^2 = a^2 + b^2 - 2(ab)\cos C$

  He must determine $B$ for the first triangle, so he can see that he must use equation $2$, which is the only one that includes the angle $B$.

  He isolates $B$ in the equation using opposite operations:

  $b^2 = a^2 + c^2 - 2(ac)\cos B$

  $2(ac)\cos B = a^2 + c^2 - b^2$

  $\cos B = \dfrac{a^2 + c^2 - b^2}{2(ac)}$

  He finishes isolating $B$ by applying $\cos^{-1}$ to both sides:

  $B = \cos^{-1}(\dfrac{a^2 + c^2 - b^2}{2(ac)})$

  He knows that the side lengths for the first triangle are:

  $a = 11$

  $b = 12$

  $c = 8$

  Substituting in these values, he gets:

  $B = \cos^{-1}(\dfrac{11^2 + 8^2 - 12^2}{2(11)(8)})$

  He finds angle $B$ to be $76.5°$.

  Similarly, for the second triangle, he uses equation $3$, and finds angle $C$ to be $27.7°$.

  For the second triangle, he uses equation $1$, and finds angle $A$ to be $112.6°$.