Example 1
Completion requirements
| Example 1 |
Solve for side \(g\) in the following triangle. Round to the nearest tenth of a millimetre.
Step 1: Determine the height of the triangle.
\(\theta = 60^\circ\)
\({\mathop{\rm opp}\nolimits} = h\)
\({\mathop{\rm hyp}\nolimits} = 17\thinspace \rm{mm}\)
\(\begin{align}
\sin \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\sin 60^\circ &= \frac{h}{{17}} \\
17\sin 60^\circ &= h \\
14.722... &= h \\
\end{align}\)
Step 2: Determine the length of side \(g\).
Remember not to round until the final step!The length of side \(g\) is approximately \(84.8 \thinspace \rm{mm}\)
\(\theta = 60^\circ\)
\({\mathop{\rm opp}\nolimits} = h\)
\({\mathop{\rm hyp}\nolimits} = 17\thinspace \rm{mm}\)
\(\begin{align}
\sin \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\sin 60^\circ &= \frac{h}{{17}} \\
17\sin 60^\circ &= h \\
14.722... &= h \\
\end{align}\)
Step 2: Determine the length of side \(g\).
\(\theta = 10^\circ\)
\({\mathop{\rm opp}\nolimits} = h = 14.722...\)
\({\mathop{\rm hyp}\nolimits} = g\)
\(\begin{align}
\sin \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\sin 10^\circ &= \frac{{14.722...}}{g} \\
g &= \frac{{14.722...}}{{\sin 10^\circ }} \\
g &= 84.783... \\
g &\doteq 84.8 \\
\end{align}\)
\({\mathop{\rm opp}\nolimits} = h = 14.722...\)
\({\mathop{\rm hyp}\nolimits} = g\)
\(\begin{align}
\sin \theta &= \frac{{{\mathop{\rm opp}\nolimits} }}{{{\mathop{\rm hyp}\nolimits} }} \\
\sin 10^\circ &= \frac{{14.722...}}{g} \\
g &= \frac{{14.722...}}{{\sin 10^\circ }} \\
g &= 84.783... \\
g &\doteq 84.8 \\
\end{align}\)
Remember not to round until the final step!