Use the law of sines and/or the law of cosines to solve non-right triangles.
Subsection8.4.1Activities
Remark8.4.1.
In Section 6.3 and Section 6.4, we learned how to solve right triangles (and special right triangles) using trigonometric ratios. In this section, we will learn how to solve oblique (non-right triangles).
Activity8.4.2.
Suppose you are given the following triangle, where \(h\) is the altitude of triangle \(ABC\text{.}\)
Figure8.4.3.
(a)
Find \(\sin A\text{.}\)
\(\displaystyle \sin A=\frac{a}{h}\)
\(\displaystyle \sin A=\frac{h}{b}\)
\(\displaystyle \sin A=\frac{h}{a}\)
\(\displaystyle \sin A=\frac{a}{b}\)
Answer.
B
(b)
Find \(\sin B\text{.}\)
\(\displaystyle \sin B=\frac{a}{h}\)
\(\displaystyle \sin B=\frac{h}{b}\)
\(\displaystyle \sin B=\frac{h}{a}\)
\(\displaystyle \sin B=\frac{a}{b}\)
Answer.
C
(c)
Take your answers from parts (a) and (b) and solve each equation for \(h\) (the altitude). What does \(h\) equal in each case?
Answer.
\(h=b \cdot \sin A\)
\(h=a \cdot \sin B\)
(d)
Notice that we now have two ways to express \(h\text{.}\) Set the two expressions of \(h\) equal to one another and then rearrange this equation so that \(a\) and \(\sin A\) are on the same side.
Answer.
\(\frac{\sin A}{a}=\frac{\sin B}{b}\) or \(\frac{a}{\sin A}=\frac{b}{\sin B}\)
Theorem8.4.4.
The Law of Sines defines the ratio of sides of a triangle and their respective sine angles are equivalent to each other.
When using the Law of Sines, you will only need to set one fraction to one other fraction. Note that these are proportions and so the reciprocals also hold true.
Remark8.4.5.
The Law of Sines is helpful when we have
two angles and one of their corresponding sides.
two sides and one of their corresponding angles.
Activity8.4.6.
Suppose you are given triangle \(ABC\) where \(A=35\)°, \(B=25\)°, and \(a=10\text{.}\)
(a)
Draw triangle \(ABC\) and label what is known. What parts of the triangle are missing?
Answer.
Students should be able to draw the triangle and label the appropriate sides/angles. The following parts of the triangle are missing: \(C\text{,}\)\(b\text{,}\) and \(c\text{.}\)
(b)
What is the value of \(C\text{?}\)
\(\displaystyle 60^\circ\)
\(\displaystyle 155^\circ\)
\(\displaystyle 145^\circ\)
\(\displaystyle 120^\circ\)
Answer.
D
(c)
Use the Law of Sines to set up a proportion to find \(c\text{.}\) What does that proportion look like?
Answer.
\(\frac{\sin 35^\circ}{10}=\frac{\sin 120^\circ}{c}\) or \(\frac{10}{\sin 35^\circ}=\frac{c}{\sin 120^\circ}\)
(d)
Solve the proportion you got in part (c) and solve for \(c\text{.}\) What is the value of \(c\) to the nearest tenth?
\(\displaystyle 8.7\)
\(\displaystyle 6.6\)
\(\displaystyle 15.1\)
\(\displaystyle 0.1\)
Answer.
C
(e)
We now have side \(b\) left to find. Which method would be the best method to find \(b\text{?}\)
Use the Law of Sines again using side \(c\text{.}\)
Pythagorean Theorem
Use the Law of Sines again with angle \(B\text{.}\)
Answer.
A and C
(f)
Use the method you identified in the previous part to find \(b\text{.}\) What is the value of \(b\) to the nearest tenth?
\(\displaystyle 5.7\)
\(\displaystyle 4.2\)
\(\displaystyle 13.6\)
\(\displaystyle 7.4\)
Answer.
D
Remark8.4.7.
The triangle you just drew in Activity 8.4.6 is known as an AAS triangle, which means we know two angles and one side (which is NOT between the angles). This is just one type of triangle you might see when solving oblique triangles. The next activity will show another type of triangle you might encounter.
Activity8.4.8.
Suppose you are given triangle \(ABC\) where \(A=76\)°, \(B=34\)°, and \(c=9\text{.}\)
(a)
Draw triangle \(ABC\) and label what is known. What parts of the triangle are missing?
Answer.
Students should draw the triangle and label the appropriate sides/angles. The following parts of the triangle are missing: \(C\text{,}\)\(a\text{,}\) and \(b\text{.}\)
(b)
What is the value of \(C\text{?}\)
\(70\)°
\(110\)°
\(104\)°
\(146\)°
Answer.
A
(c)
Use the Law of Sines to set up a proportion to find \(a\text{.}\) What is the value of \(a\) to the nearest tenth?
\(\displaystyle 8.7\)
\(\displaystyle 9.3\)
\(\displaystyle 8.5\)
\(\displaystyle 9.1\)
Answer.
(d)
Use the Law of Sines to set up a proportion to find \(b\text{.}\) What is the value of \(b\) to the nearest tenth?
\(\displaystyle 5.0\)
\(\displaystyle 8.5\)
\(\displaystyle 15.1\)
\(\displaystyle 5.4\)
Answer.
Remark8.4.9.
The triangle you just drew in Activity 8.4.8 is known as an ASA triangle, which means we know two angles and a side BETWEEN the angles.
Activity8.4.10.
Suppose you are given triangle \(ABC\text{,}\) where \(A = 30\)°, \(a = 7\text{,}\) and \(b = 16\text{.}\)
(a)
Draw triangle \(ABC\) and label what is known. What parts of the triangle are missing?
Answer.
Students should draw the triangle and label the appropriate sides/angles. The following parts of the triangle are missing: \(B\text{,}\)\(C\text{,}\) and \(c\text{.}\)
(b)
What is the value of \(\sin B\text{?}\)
\(\displaystyle 8.0\)
\(\displaystyle 1.14\)
\(\displaystyle 0.875\)
\(\displaystyle 0.031\)
Answer.
B
(c)
Use part (b) to determine the value of \(B\text{.}\)
Answer.
Students will see that when finding \(\sin^{-1}(\frac{8}{7})\text{,}\) that the value does not exist.
Remark8.4.11.
In Activity 8.4.10, we can see that no triangle can be created because \(\sin B\) was equal to a value greater than \(1\text{.}\) This is because the largest value that the sine of an angle can have is \(1\) (refer back to Section 6.5).
Activity8.4.12.
Suppose you are given triangle \(ABC\text{,}\) where \(A = 30\)°, \(a = 10\text{,}\) and \(b = 16\text{.}\)
(a)
Use the Law of Sines to determine the value of \(B\) (to the nearest degree).
\(143\)°
\(53\)°
\(80\)°
\(37\)°
Answer.
B
(b)
Continue solving triangle \(ABC\text{.}\) What are the missing values (to the nearest degree/whole number)?
Answer.
\(C\approx 97\)°
\(c \approx 20\)
(c)
Let’s go back to part (a) when we were asked to solve for \(B\text{.}\) There is another value of \(B\) that exists. Which of the following values could also be \(B\text{?}\)
\(143\)°
\(37\)°
\(80\)°
\(127\)°
Answer.
D
(d)
Why do you think there are two values of \(B\) for this triangle?
Answer.
Because \(\sin \theta\) exists in both Quadrant I and Quadrant II, there exists two possible values of \(\theta\text{.}\)\(\sin 30\)° and \(\sin 150\)°, for example, are both equal to \(\frac{1}{2}\text{.}\)
(e)
Continue solving this other triangle \(ABC\text{.}\) What are the missing values (to the nearest degree/whole number)?
Answer.
\(C \approx 23\)°
\(c \approx 8\)
Remark8.4.13.
So far we have seen that when given two sides and an angle (also known as a SSA triangle), we can have no solution (i.e., no triangle can be created) or two solutions (i.e., there are two possible triangles). There is still one more case we need to explore.
Activity8.4.14.
Suppose you are given triangle \(ABC\text{,}\) where \(A = 30\)°, \(a = 20\text{,}\) and \(b = 16\text{.}\)
(a)
Use the Law of Sines to determine the value of \(B\) (to the nearest degree).
\(24\)°
\(39\)°
\(141\)°
\(156\)°
Answer.
A and D
(b)
Choose the smaller value of \(B\) and solve triangle \(ABC\text{.}\) What are the missing values (to the nearest degree/whole number)?
Answer.
\(C \approx 126\)°
\(c \approx 32\)
(c)
Choose the larger value of \(B\) and solve triangle \(ABC\text{.}\) What are the missing values (to the nearest degree/whole number)?
Answer.
Another triangle does not exist because if \(B\) is equal to \(156\)°, then that would mean the third angle, \(C\text{,}\) would have to be \(-6\)°.
Observation8.4.15.
In the previous three activities, we saw that when we use the Law of Sines to find an angle, an ambiguity can arise due to the sine function being positive in Quadrant I and Quadrant II. In other words, if two sides and the non-included angle are given (SSA), three situations may occur.
NO triangle exists (no solution)
TWO different triangle exist (2 solutions)
Exactly ONE triangle exists (1 solution)
The Ambiguous Case of the Law of Sines states that when using the Law of Sines to find a missing side length, the possibility of two solutions for the measure of the same side may occur.
Activity8.4.16.
State the number of possible triangles that can be formed with the given measurements. Then, solve each triangle. Round your answers to the nearest tenth.
(a)
\(C=145\)°, \(b=7\text{,}\)\(c=33\)
Answer.
One triangle: \(A=28\)°, \(B=7\)°, \(a=27\)
(b)
\(B=84\)°, \(a=18\text{,}\)\(b=9\)
Answer.
Not a triangle
(c)
\(B=45\)°, \(a=28\text{,}\)\(b=27\)
Answer.
Two triangles exist:
\(C=87.8\)°, \(A=47.2\)°, \(c=38.2\)
\(C=2.2\)°, \(A=132.8\)°, \(c=1.5\)
Activity8.4.17.
Suppose you are given triangle \(ABC\text{,}\) where \(A = 70\)°, \(b = 14\text{,}\) and \(c = 7\text{.}\)
(a)
Draw triangle \(ABC\) and label what is known. What parts of the triangle are missing?
Answer.
Students should draw the triangle and label the appropriate sides/angles. The following parts of the triangle are missing: \(B\text{,}\)\(C\text{,}\) and \(a\text{.}\)
(b)
Use the Law of Sines to solve triangle \(ABC\text{.}\)
Answer.
Students should recognize that the Law of Sines cannot be used to solve this triangle.
Remark8.4.18.
Notice in Activity 8.4.17, we do not currently have enough information to be able to solve for triangle \(ABC\) since the Law of Sines cannot be used. In the next activity, we will explore another method that can be used to solve oblique triangles.
Activity8.4.19.
Suppose you are given the following triangle, where \(h\) is the altitude of triangle \(ABC\text{.}\)
Figure8.4.20.
(a)
Using right triangle trigonometry, find \(\sin A\text{.}\)
\(\displaystyle \sin A=\frac{b}{h}\)
\(\displaystyle \sin A=\frac{h}{b}\)
\(\displaystyle \sin A=\frac{m}{b}\)
\(\displaystyle \sin A=\frac{b}{m}\)
Answer.
B
(b)
Using right triangle trigonometry, find \(\cos A\text{.}\)
\(\displaystyle \cos A=\frac{b}{h}\)
\(\displaystyle \cos A=\frac{h}{b}\)
\(\displaystyle \cos A=\frac{m}{b}\)
\(\displaystyle \cos A=\frac{b}{m}\)
Answer.
C
(c)
Take your equation in part (a) and solve for \(h\text{.}\)
Answer.
\(h=b \sin A\)
(d)
Take your equation in part (b) and solve for \(m\text{.}\)
Answer.
\(m=b \cos A\)
(e)
Recall that we can use the Pythagorean Theorem to represent the relationship between the sides of a right triangle. For example, \(h^2+ (c-m)^2 = a^2\) can be used to represent the relationship of the sides of triangle \(BCD\text{.}\) Take your equations from parts (c) and (d) to substitute \(m\) and \(h\) into that equation.
Hint.
When simplifying the term \((c-b \cos A)^2\text{,}\) it might be easier to multiply \((c-b \cos A)*(c-b \cos A)\text{.}\)
Answer.
\(h^2+ (c-m)^2 = a^2\)
\((b \sin A)^2+ (c-(b \cos A))^2=a^2\)
(f)
Simplify your equation from part (e).
Hint.
You may have to use a trig identity to simplify!
Answer.
\((b \sin A)^2+ (c-(b \cos A))^2=a^2\)
\(b^2 \sin^2 A + c^2 -2bc \cos A + b^2 \cos^2 A = a^2\)
\(b^2 \sin^2 A + b^2 \cos^2 A + c^2 -2bc \cos A= a^2\)
\(b^2 (\sin^2 A + \cos^2 A) + c^2 -2bc \cos A= a^2\)
\(b^2 + c^2 -2bc \cos A= a^2\)
Theorem8.4.21.
The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. In other words, the square of any one side of a triangle is equal to the difference between the sum of squares of the other two sides and double the product of other sides and cosine angle included between them.
Let \(a\text{,}\)\(b\text{,}\) and \(c\) be the lengths of the three sides of a triangle and \(A\text{,}\)\(B\text{,}\) and \(C\) be the three angles of the triangle. Then, the Law of Cosines states that:
\begin{equation*}
a^2=b^2+c^2-2bc\cos A
\end{equation*}
\begin{equation*}
b^2=a^2+c^2-2ac\cos B
\end{equation*}
\begin{equation*}
c^2=a^2+b^2-2ab\cos C
\end{equation*}
Activity8.4.22.
Let’s revisit Activity 8.4.17 to see how we can apply the Law of Cosines to solve for this triangle. Recall that we were given triangle \(ABC\text{,}\) where \(A = 70\)°, \(b = 14\text{,}\) and \(c = 7\text{.}\)
(a)
Draw triangle \(ABC\) and label what is known. What parts of the triangle are missing?
Answer.
Students should draw the triangle and label the appropriate sides/angles. The following parts of the triangle are missing: \(B\text{,}\)\(C\text{,}\) and \(a\text{.}\)
(b)
Use the Law of Cosines to find \(a\) (to the nearest tenth).
\(\displaystyle 178.0\)
\(\displaystyle 17.7\)
\(\displaystyle 13.3\)
\(\displaystyle 7.8\)
Answer.
C
(c)
Now that we have another side of triangle \(ABC\text{,}\) what would you use to find angle \(B\text{?}\)
Law of Sines
Law of Cosines
Pythagorean Theorem
Answer.
A or B. Students should discuss how either method can used, but that there is a possibility of the ambiguous case for the Law of Sines.
(d)
Find the measure of \(B\) and \(C\) (to the nearest degree).
Answer.
\(B \approx 82\)° and \(C \approx 28\)°
Activity8.4.23.
Suppose you are given triangle \(ABC\text{,}\) where \(a = 30\text{,}\)\(b = 20\text{,}\) and \(c = 17\text{.}\)
(a)
Draw triangle \(ABC\) and label what is known. What parts of the triangle are missing?
Answer.
Students should draw the triangle and label the appropriate sides/angles. The following parts of the triangle are missing: \(A\text{,}\)\(B\text{,}\) and \(C\text{.}\)
(b)
Which law should you start with to solve triangle \(ABC\text{?}\)
Answer.
Law of Cosines
(c)
What is the value of \(A\) (to the nearest degree)?
\(109\)°
\(39\)°
\(108\)°
\(33\)°
Answer.
C
(d)
Now that you have found angle \(A\text{,}\) which law would be the best to use to find the measure of angle \(B\text{?}\)
Answer.
The Law of Sines might be the easiest method to use, but students could also use the Law of Cosines.
(e)
What is the value of \(B\) (to the nearest degree)?
\(109\)°
\(39\)°
\(108\)°
\(33\)°
Answer.
B
(f)
What is the value of \(C\) (to the nearest degree)?
\(109\)°
\(39\)°
\(108\)°
\(33\)°
Answer.
D
Remark8.4.24.
Notice in Activity 8.4.23, you were given a triangle with three known sides. This type of triangle is known as a side-side-side (SSS) triangle.
Activity8.4.25.
State whether the Law of Sines or the Law of Cosines is the best choice to use to determine the indicated angle/side.
Hint.
It might help to draw a picture!
(a)
Given \(B=112\)°, \(a=12\text{,}\) and \(b=25\text{,}\) find \(A\text{.}\)
Answer.
Law of Sines
(b)
Given \(B=87\)°, \(a=15\text{,}\) and \(c=16\text{,}\) find \(b\text{.}\)
Answer.
Law of Cosines
(c)
Given \(a=37\text{,}\)\(b=55\text{,}\) and \(c=30\text{,}\) find \(C\text{.}\)
Answer.
Law of Cosines
(d)
Given \(A=108\)°, \(B=40\)°, and \(b=20\text{,}\) find \(c\text{.}\)
Answer.
Law of Sines
Activity8.4.26.
Now that we can determine which laws to use, let’s go back to Activity 8.4.25 to solve each triangle (to the nearest degree/whole number).
(a)
Given \(B=112\)°, \(a=12\text{,}\) and \(b=25\text{,}\) find \(A\text{.}\)
Answer.
\(A \approx 26\)°, \(C \approx 42\)°, and \(c \approx 18\)
(b)
Given \(B=87\)°, \(a=15\text{,}\) and \(c=16\text{,}\) find \(b\text{.}\)
Answer.
\(b \approx 21\text{,}\)\(A \approx 46\)°, and \(C \approx 47\)°
(c)
Given \(a=37\text{,}\)\(b=55\text{,}\) and \(c=30\text{,}\) find \(C\text{.}\)
Answer.
\(C \approx 31\)°, \(A \approx 39\)°, and \(B \approx 110\)°
(d)
Given \(A=108\)°, \(B=40\)°, and \(b=20\text{,}\) find \(c\text{.}\)
Answer.
\(C \approx 38\)°, \(b \approx 82\text{,}\) and \(c \approx 52\)
