In Section 6.4 and Section 8.4, 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).
Suppose you are given the following triangle, where \(h\) is the altitude of triangle \(ABC\text{.}\) By drawing the altitude, we’ve now created two right triangles. If we label the point where the altitude intersects \(C\) as point \(D\text{,}\) then we can call the triangles \(ACD\) and \(BCD\text{.}\)
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.
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.
The triangle in Activity 8.5.6 is known as an \(AAS\) (Angle-Angle-Side) triangle, which means we know two angles and one side (which is NOT between the angles) in that order. 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.
In Activity 8.5.12, 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).
Let’s go back to part (b) when we were asked to solve for \(B\text{.}\) We needed to solve \(\sin B=\frac{4}{5}\text{.}\) Using the inverse sine function on our calculator, we got that \(B \approx 53\)°. However, there is another angle between \(0^\circ\) and \(180^\circ\) whose sine is \(\frac{4}{5}\text{.}\) Which of the following values could also be \(B\text{?}\)
So far we have seen that when given two sides and an angle (also known as a \(SSA\) (Side-Side-Angle) 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.
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.
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.
Figure8.5.19.If \(a\) is too short for a given angle \(A\) and side \(b\text{,}\) no matter how that leg is swung around the dashed circle, it will not meet a third side of any length along the dotted line.
Figure8.5.20.If \(a\) is larger, the circle of radius \(a\) centered at \(C\) intersects the horizontal dashed line in two places, giving two possible solutions.
Figure8.5.21.If the length of \(a\) is the same as the height of the triangle, the two possibilities converge and there is a single solution of a right triangle. On the other hand, if the length of \(a\) increases enough, the potential second solution swings past the angle \(A\) and thus does not exist.
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.
Notice in Activity 8.5.23, 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.
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.
When simplifying the term \((c-b \cos A)^2\text{,}\) don’t forget that \((c-b \cos A)^2=(c-b \cos A)*(c-b \cos A)\text{.}\) You may also have to use a trig identity to simplify!
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*}
Let’s revisit Activity 8.5.23 to see how we can apply the Law of Cosines to solve for this triangle. Recall that we were given triangle \(ABC\text{,}\) where \(A = 70^\circ\text{,}\)\(b = 14\text{,}\) and \(c = 7\text{.}\)
