Solving General Triangles
Solved Problems
Example 1.
Solve the triangle \(ABC\) in which \(c = 50,\) \(\alpha = 60^\circ,\) and \(\beta = 110^\circ.\)
Solution.
This is an example of the \(ASA\) case, so first we find the unknown angle \(\gamma:\)
\[\gamma = 180^\circ - \alpha - \beta = 180^\circ - 60^\circ - 110^\circ = 10^\circ .\]
Now using the Sine Rule, we can calculate the sides \(a\) and \(b:\)
\[\frac{a}{{\sin \alpha }} = \frac{c}{{\sin \gamma }}, \Rightarrow a = \frac{{c\sin \alpha }}{{\sin \gamma }} = \frac{{c\sin 60^\circ }}{{\sin 10^\circ }} = \frac{{50 \times 0,8660}}{{0,1736}} = 249,36 \approx 249;\]
\[\frac{b}{{\sin \beta }} = \frac{c}{{\sin \gamma }}, \Rightarrow b = \frac{{c\sin \beta }}{{\sin \gamma }} = \frac{{c\sin 110^\circ }}{{\sin 10^\circ }} = \frac{{50 \times 0,9397}}{{0,1736}} = 270,65 \approx 271.\]
Example 2.
Solve the triangle \(ABC\) if \(b = 12\,\text{cm},\) \(c = 6\,\text{cm},\) and \(\alpha = 60^\circ.\)
Solution.
We know two sides and the angle between them, that is, we have the \(SAS\) case. Therefore, we will use the Law of Cosines to solve the triangle.
Find the side \(a:\)
\[{a^2} = {b^2} + {c^2} - 2bc\cos \alpha , \Rightarrow {a^2} = {12^2} + {6^2} - 2 \cdot 12 \cdot 6 \cdot \cos 60^\circ = 144 + 36 - 144 \cdot \frac{1}{2} = 108, \Rightarrow a = \sqrt {108} = 6\sqrt 3\,\text{cm}.\]
The angle \(\beta\) can also be found by the Cosine Rule:
\[\cos \beta = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} = \frac{{{{\left( {6\sqrt 3 } \right)}^2} + {6^2} - {{12}^2}}}{{2 \cdot 6\sqrt 3 \cdot 6}} = \frac{{\cancel{{108}} + \cancel{{36}} - \cancel{{144}}}}{{72\sqrt 3 }} = 0.\]
We see that \(\beta = 90^\circ,\) so \(\triangle ABC\) is a right triangle.
The remaining angle \(\gamma\) is equal to
\[\gamma = 180^\circ - \alpha - \beta = 180^\circ - 60^\circ - 90^\circ = 30^\circ .\]
Example 3.
Solve the triangle \(ABC\) with sides \(a = 15,\) \(b = 25,\) \(c = 35.\)
Solution.
We have here the \(SSS\) case. Find the angle \(\alpha\) by the Cosine Rule:
\[\cos \alpha = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{{{25}^2} + {{35}^2} - {{15}^2}}}{{2 \cdot 25 \cdot 35}} = \frac{{625 + 1225 - 225}}{{1750}} = \frac{{1625}}{{1750}} = \frac{{13}}{{14}}.\]
Then
\[\alpha = \arccos \frac{{13}}{{14}} \approx 21.8^\circ .\]
Similarly we calculate the angle \(\beta:\)
\[\cos \beta = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} = \frac{{{{15}^2} + {{35}^2} - {{25}^2}}}{{2 \cdot 15 \cdot 35}} = \frac{{225 + 1225 - 625}}{{1050}} = \frac{{825}}{{1050}} = \frac{{11}}{{14}}.\]
Hence,
\[\beta = \arccos \frac{{11}}{{14}} \approx 38.2^\circ .\]
The angle \(\gamma\) is given by
\[\gamma = 180^\circ - \alpha - \beta = 180^\circ - 21.8^\circ - 38.2^\circ = 120^\circ .\]
Example 4.
Solve the triangle \(ABC\) in which \(a = 12\,\text{cm},\) \(b = 5\,\text{cm},\) and \(\alpha = 120^\circ.\)
Solution.
We are given an \(SSA\) triangle, which includes two sides and the angle opposite one of these sides. This set of parameters is classified as an ambiguous case since it may have more than one solution. The unique solution exists if only the angle is opposite the longest side of the two given sides. In our case \(b \le a,\) so the triangle is well defined.
Figure 6. Using the Law of Sines, we find the angle \(\beta:\)
\[\frac{b}{{\sin \beta }} = \frac{a}{{\sin \alpha }}, \Rightarrow \sin \beta = \frac{{b\sin \alpha }}{a} = \frac{{5\sin 120^\circ }}{{12}} = \frac{{5 \cdot \frac{{\sqrt 3 }}{2}}}{{12}} = \frac{{5\sqrt 3 }}{{24}}.\]
It follows from here that
\[\beta = \arcsin \frac{{5\sqrt 3 }}{{24}} \approx 21,2^\circ .\]
Determine the third angle \(\gamma:\)
\[\gamma = 180^\circ - \alpha - \beta = 180^\circ - 120^\circ - 21,2^\circ = 38,8^\circ .\]
The side \(c\) can be found from the Law of Sines:
\[\frac{a}{{\sin \alpha }} = \frac{c}{{\sin \gamma }}, \Rightarrow c = \frac{{a\sin \gamma }}{{\sin \alpha }} = \frac{{12\sin 38,8^\circ }}{{\sin 120^\circ }} = \frac{{12 \cdot 0,6267}}{{\frac{{\sqrt 3 }}{2}}} = 8,68 \approx 8,7\,\text{cm}.\]
Example 5.
Given a parallelogram with sides \(a\) and \(b.\) Find the length of the longer diagonal \(d_2,\) if the length of the shorter diagonal is \(d_1.\)
Solution.
Figure 7. Let the angle \(C\) be equal to \(\alpha.\) Write the Law of Cosines for triangle \(BCD:\)
\[d_1^2 = {a^2} + {b^2} - 2ab\cos \alpha \]
and express \(\cos \alpha\) in terms of \(a, b, d_1:\)
\[\cos \alpha = \frac{{{a^2} + {b^2} - d_1^2}}{{2ab}}.\]
Similarly, given that \(\angle B = 180^\circ - \alpha\), write the Cosine Rule for \(\triangle ABC:\)
\[d_2^2 = {a^2} + {b^2} - 2ab\cos \left( {180^\circ - \alpha } \right).\]
By the reduction identity , \(\cos \left( {180^\circ - \alpha } \right) = - \cos \alpha .\) Hence,
\[d_2^2 = {a^2} + {b^2} - 2ab\left( { - \cos \alpha } \right) = {a^2} + {b^2} + 2ab\cos \alpha .\]
Substitute \(\cos \alpha\) from the previous equation to obtain the length of the longer diagonal \(d_2:\)
\[d_2^2 = {a^2} + {b^2} + 2ab \cdot \frac{{{a^2} + {b^2} - d_1^2}}{{2ab}} = {a^2} + {b^2} + {a^2} + {b^2} - d_1^2 = 2\left( {{a^2} + {b^2}} \right) - d_1^2.\]
Hence,
\[{d_2} = \sqrt {2\left( {{a^2} + {b^2}} \right) - d_1^2} .\]
Example 6.
Derive Mollweide's formula \[\frac{{a + b}}{c} = \frac{{\cos \frac{{\alpha - \beta }}{2}}}{{\sin \frac{\gamma }{2}}},\] where \(a,b,c\) are the sides of an oblique triangle, and \(\alpha, \beta, \gamma\) are the angles opposite to these sides, respectively.
Solution.
To prove the formula, we write the Sine Rule for the triangle:
\[\frac{a}{{\sin \alpha }} = \frac{b}{{\sin \beta }} = \frac{c}{{\sin \gamma }}.\]
It follows from this relationship that
\[\frac{a}{c} = \frac{{\sin \alpha }}{{\sin \gamma }} \;\;\text{and}\;\;\frac{b}{c} = \frac{{\sin \beta }}{{\sin \gamma }}.\]
Adding the two last equations gives us
\[\frac{a}{c} + \frac{b}{c} = \frac{{\sin \alpha }}{{\sin \gamma }} + \frac{{\sin \beta }}{{\sin \gamma }} = \frac{{\sin \alpha + \sin \beta }}{{\sin \gamma }}.\]
Now we transform the expression in the numerator using the sum-to-product identity :
\[\sin \alpha + \sin \beta = 2\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}.\]
We also apply the double angle formula to \(\sin \gamma\) in the denominator. As a result, we get
\[\frac{{a + b}}{c} = \frac{{\sin \alpha + \sin \beta }}{{\sin \gamma }} = \frac{{2\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}}}{{2\sin \frac{\gamma }{2}\cos \frac{\gamma }{2}}}.\]
Since \(\alpha + \beta + \gamma = 180^\circ,\) then using the cofunction identity , we have
\[\sin \frac{{\alpha + \beta }}{2} = \sin \frac{{180^\circ - \gamma }}{2} = \cos \left( {90^\circ - \frac{\gamma }{2}} \right) = \sin \frac{\gamma }{2},\]
that is,
\[\frac{{a + b}}{c} = \frac{{\cancel{2}\cancel{{\cos \frac{\gamma }{2}}}\cos \frac{{\alpha - \beta }}{2}}}{{\cancel{2}\sin \frac{\gamma }{2}\cancel{{\cos \frac{\gamma }{2}}}}} = \frac{{\cos \frac{{\alpha - \beta }}{2}}}{{\sin \frac{\gamma }{2}}}.\]
Thus, we have proved Mollweide's formula
\[\frac{a + b}{c} = \frac{\cos\frac{\alpha - \beta}{2}}{\sin\frac{\gamma}{2}}\]
