B. The Cosine Law
Completion requirements
B. The Cosine Law
In Section A, you determined that solving for a side or an angle in a non-right triangle, using the basic trigonometric ratios, can be accomplished by splitting the triangle into two right triangles. Once you have two right triangles, you are able to find a solution, but it does require many steps. Can a general rule be derived that will allow you to solve for a side of an oblique triangle in one step?
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Consider the non-right triangle shown. Determine the length of side \(a\).
The side and angle opposite each other are given the same letter. Sides are given the lowercase letter and angles are given the uppercase letter. For example, side \(a\) is opposite angle \(A\).
First, split the triangle into two right triangles by drawing an altitude segment from vertex \(B\), intersecting side \(AC\) at a right angle, at point \(D\).
The resulting triangles share side \(h\). Triangle \(BAD\) has a base of \(x\), and triangle \(BCD\) has a base of \(b - x\). In order to determine the length of \(a\), you need to determine expressions for \(h\) and \(x\), using triangle \(BAD\).
Using the Pythagorean Theorem on triangles \(BCD\) and \(BAD\), you have \(a^2 = h^2 + \left( {b - x} \right)^2 \) and \(c^2 = h^2 + x^2 \), respectively. Notice that the second equation can be rearranged so that \(c^2 - x^2 = h^2 \). Substituting the rearranged equation for \(h_2\) in \(a^2 = h^2 + \left( {b - x} \right)^2 \) gives \(a^2 = c^2 - x^2 + \left( {b - x} \right)^2\).
Expand and simplify the resulting equation.
\(\begin{align}
a^2 &= c^2 - x^2 + \left( {b - x} \right)\left( {b - x} \right) \\
a^2 &= c^2 - x^2 + b^2 - 2bx + x^2 \\
a^2 &= c^2 + b^2 - 2bx \\
a^2 &= b^2 + c^2 - 2bx \\
\end{align}\)
In order to eliminate the \(x\) variable, which was only introduced when the altitude was drawn, determine an expression for \(x\), using the cosine of angle \(A\) in triangle \(BAD\).
\(\begin{align}
\cos \theta &= \frac{{{\mathop{\rm adj}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\cos A &= \frac{x}{c} \\
c\cos A &= x \\
\end{align}\)
Now, substitute \(x = c\cos A\) for \(x\) in \(a^2 = b^2 + c^2 - 2bx\).
\(a^2 = b^2 + c^2 - 2bc\cos A\)
You now have a formula that will help you solve for a side of an oblique triangle. The formula \(a^2 = b^2 + c^2 - 2bc\cos A\) is called the cosine law.
The cosine law can also be rearranged to help you solve for an angle.
\(\begin{align}
a^2 &= b^2 + c^2 - 2bc\cos A \\
2bc\cos A &= b^2 + c^2 - a^2 \\
\cos A &= \frac{{b^2 + c^2 - a^2 }}{{2bc}} \\
\end{align}\)
The side and angle opposite each other are given the same letter. Sides are given the lowercase letter and angles are given the uppercase letter. For example, side \(a\) is opposite angle \(A\).
First, split the triangle into two right triangles by drawing an altitude segment from vertex \(B\), intersecting side \(AC\) at a right angle, at point \(D\).
The resulting triangles share side \(h\). Triangle \(BAD\) has a base of \(x\), and triangle \(BCD\) has a base of \(b - x\). In order to determine the length of \(a\), you need to determine expressions for \(h\) and \(x\), using triangle \(BAD\).
Using the Pythagorean Theorem on triangles \(BCD\) and \(BAD\), you have \(a^2 = h^2 + \left( {b - x} \right)^2 \) and \(c^2 = h^2 + x^2 \), respectively. Notice that the second equation can be rearranged so that \(c^2 - x^2 = h^2 \). Substituting the rearranged equation for \(h_2\) in \(a^2 = h^2 + \left( {b - x} \right)^2 \) gives \(a^2 = c^2 - x^2 + \left( {b - x} \right)^2\).
Expand and simplify the resulting equation.
\(\begin{align}
a^2 &= c^2 - x^2 + \left( {b - x} \right)\left( {b - x} \right) \\
a^2 &= c^2 - x^2 + b^2 - 2bx + x^2 \\
a^2 &= c^2 + b^2 - 2bx \\
a^2 &= b^2 + c^2 - 2bx \\
\end{align}\)
In order to eliminate the \(x\) variable, which was only introduced when the altitude was drawn, determine an expression for \(x\), using the cosine of angle \(A\) in triangle \(BAD\).
\(\begin{align}
\cos \theta &= \frac{{{\mathop{\rm adj}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\cos A &= \frac{x}{c} \\
c\cos A &= x \\
\end{align}\)
Now, substitute \(x = c\cos A\) for \(x\) in \(a^2 = b^2 + c^2 - 2bx\).
\(a^2 = b^2 + c^2 - 2bc\cos A\)
You now have a formula that will help you solve for a side of an oblique triangle. The formula \(a^2 = b^2 + c^2 - 2bc\cos A\) is called the cosine law.
The cosine law can also be rearranged to help you solve for an angle.
\(\begin{align}
a^2 &= b^2 + c^2 - 2bc\cos A \\
2bc\cos A &= b^2 + c^2 - a^2 \\
\cos A &= \frac{{b^2 + c^2 - a^2 }}{{2bc}} \\
\end{align}\)