Sine Law

Introduction:

In trigonometry, the law of sinessine lawsine formula, or sine rule is an equation relating the lengths of the sides of a triangle (any shape) to the sines of its angles.

The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the triangle is not uniquely determined by this data (called the ambiguous case) and the technique gives two possible values for the enclosed angle.

The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in scalene triangles, with the other being the law of cosines.

It can be generalized to higher dimensions on surfaces with constant curvature. Let's describe it further.


What is the Sine Law?

The sine law says that the sides of a triangle are proportional to the sines of the angles opposite to them:

\[\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}\]

Let us outline a justification for this law:

Consider the following figure, which shows a \(\Delta {\rm{ABC}}\) which is acute-angled:

Angles' sines and triangle's sides are proportional

We note that

\[\begin{align}\sin B = \frac{{AD}}{c},\,\,\,&\sin C = \frac{{AD}}{b}\\ \Rightarrow \,\,\,c\sin B &= b\sin C\\ \Rightarrow \,\,\,\frac{{\sin B}}{b} &= \frac{{\sin C}}{c}\end{align}\]

The sine law follows from extending this.

yesChallenge 1: Can you prove that this law holds for obtuse-angled triangles as well?

⚡Tip: Draw \(\angle {\rm{A}}\) as obtuse angle of \(\Delta {\rm{ABC}}\) and drop a perpendicular \(AD\) from \(A\) to \(BC\) and then use a similar approach as above to prove it.


Solved Examples:

Example 1: If \(\Delta \) is the area of \(\Delta {\text{ABC}}\), show that

\[\Delta  = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B\]

Solution: Considering the below figure, we have:

Angles' sines and triangle's sides are proportional

\[\Delta  = \frac{1}{2} \times BC \times AD = \frac{1}{2}\left( a \right)\left( {b\sin C} \right)\]

\[ \Rightarrow \boxed{\Delta  = \frac{1}{2}ab\sin C}\]

Similarly, we can prove the other equality.


Example 2: \(ABCD\) is a trapezium such that \(AB\) and \(CD\) are parallel and \(CB\) is perpendicular to them. If \(\angle ADB = \theta \), \(BC = p\) and \(CD = q\), show that

\[AB = \frac{{\left( {{p^2} + {q^2}} \right)\sin \theta }}{{p\cos \theta  + q\sin \theta }}\]

Solution: Consider the following figure:

 Applying Sine law for trapezium

Applying the sine law to \(\Delta {\text{ABD}}\), we have:

\[\begin{align}
  \frac{{AB}}{{\sin \theta }} &= \frac{{BD}}{{\sin \left( {\theta  + \phi } \right)}} \hfill \\
   \Rightarrow \,\,\,AB &= \frac{{BD\sin \theta }}{{\sin \theta \cos \phi  + \cos \theta \sin \phi }} \hfill \\
   &= \frac{{\left( {\sqrt {{p^2} + {q^2}} } \right)\sin \theta }}{{\sin \theta \left( {\frac{q}{{\sqrt {{p^2} + {q^2}} }}} \right) + \cos \theta \left( {\frac{p}{{\sqrt {{p^2} + {q^2}} }}} \right)}} \hfill \\ 
\end{align} \]

\[ \Rightarrow \boxed{AB = \frac{{\left( {{p^2} + {q^2}} \right)\sin \theta }}{{p\cos \theta  + q\sin \theta }}}\]


Example 3: Let \(\alpha \), \(\beta \), \(\gamma \) be the lengths of the internal  bisectors of angles \(A\), \(B\), \(C\) of triangle ABC. Determine the value of

\[S = \frac{1}{\alpha }\cos \frac{A}{2} + \frac{1}{\beta }\cos \frac{B}{2} + \frac{1}{\gamma }\cos \frac{C}{2}\]

in terms of the sides \(a\), \(b\), \(c\) of the triangle.

Solution: Consider the following figure, which will enable us to find an expression for \(\alpha \):

Using double-angle formula

We have,

\[\begin{align}
  {\text{ar.}}\left( {\Delta ABD} \right) + {\text{ar.}}\left( {\Delta ACD} \right) &= {\text{ar.}}\left( {\Delta ABC} \right) \hfill \\
   \Rightarrow \,\,\,\frac{1}{2}c\alpha \sin \frac{A}{2} + \frac{1}{2}b\alpha \sin \frac{A}{2} &= \frac{1}{2}bc\sin A \hfill \\ 
\end{align} \]

Resolving further, we get,

\[\begin{align}
   \Rightarrow \,\,\,\alpha  &= \frac{{bc\sin A}}{{\left( {b + c} \right)\sin \frac{A}{2}}} \hfill \\
   &= \frac{{2bc}}{{b + c}}\cos \frac{A}{2} \hfill \\ 
\end{align} \]

\[ \Rightarrow \frac{1}{\alpha }\cos \frac{A}{2} = \frac{1}{{2b}} + \frac{1}{{2c}}\]

We have used the following double-angle formula in the second last step:

\[\sin A = 2\sin \frac{A}{2}\cos \frac{A}{2}\]

Similarly, we can evaluate:

\[\begin{align}&\frac{1}{\beta }\cos \frac{B}{2} = \frac{1}{{2c}} + \frac{1}{{2a}}\\&\frac{1}{\gamma }\cos \frac{C}{2} = \frac{1}{{2a}} + \frac{1}{{2b}}\end{align}\]

Adding the three expressions will give the required sum as

\[ \Rightarrow \boxed{S = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}\]


yesChallenge 2: The lengths of two sides of a triangle are 4 cm and 5 cm, included angle is \(30^\circ \). Find the area.

Tip: Use the formula which we have derived in example 1.


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