Examples on Semiperimeter and Half Angle Formulae

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Example - 50

Evaluate  \(\begin{align}S = (b - c)\cot \frac{A}{2} + (c - a)\cot \frac{B}{2} + (a - b)\cot \frac{C}{2}\end{align}\)  .

Solution: By the half-angle formula for  \(\begin{align}tan \frac{A}{2},\end{align}\)  we have

\[\begin{align}&\cot \frac{A}{2} = \sqrt {\frac{{s(s - a)}}{{(s - b)(s - c)}}}  \\  \,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;\;= \frac{{s(s - a)}}{{\sqrt {s(s - a)(s - b)(s - c)} }} = \frac{{s(s - a)}}{\Delta }  \\   \Rightarrow\quad   &\;\;\;\;\;\;S = \frac{{s(s - a)(b - c) + s(s - b)(c - a) + s(s - c)(a - b)}}{\Delta } \\ \,\,\,\,\, &\qquad\;\;= \frac{{s\left( {({b^2} - {c^2}) + ({c^2} - {a^2}) + ({a^2} - {b^2})} \right)}}{\Delta } \\ \,\,\,\,\, &\qquad\;\;= 0 \\ \end{align}\]

Example - 51

Prove that \(\begin{align}\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \left( {\frac{{a + b + c}}{{b + c - a}}} \right)\cot \frac{A}{2}\end{align}\)

Solution:

\[\begin{align}& \text{LHS}=\frac{s(s-a)}{\Delta }+\frac{s(s-b)}{\Delta }+\frac{s(s-c)}{\Delta }\left\{ \begin{gathered}& \text{Using}\ \text{the}\ \text{expression}\ \text{for}\cot \frac{A}{2}\  \\ & \text{from}\ \text{the}\ \text{previous}\ \text{example} \\ \end{gathered} \right\} \\  & \,\,\,\,\,\,\,\,\,\,\,\,=\frac{s(3s-(a+b+c))}{\Delta } \\  &  \,\,\,\,\,\,\,\,\,\,\,\,=\frac{{{s}^{2}}}{\Delta } \\  & \text{RHS}=\left( \frac{a+b+c}{b+c-a} \right)\cot \frac{A}{2} \\  & \,\,\,\,\,\,\,\,\,\,\,\,=\frac{2s}{2s-2a}\cdot \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} \\ & \,\,\,\,\,\,\,\,\,\,\,=\frac{{{s}^{2}}}{\sqrt{s(s-a)(s-b)(s-c)}} \\  & \,\,\,\,\,\,\,\,\,\,\,=\frac{{{s}^{2}}}{\Delta } \\ \end{align}\]

Example - 52

Consider the following statements concerning \(\Delta ABC\)  :

I. The sides a, b, c and \(\Delta \) are rational

II. \(a,\;\tan \frac{B}{2},\;\tan \frac{C}{2}\) are rational

III. \(a,\;\sin A,\;\sin B,\;\sin C\) are rational

Show that I \( \Rightarrow \)  II \( \Rightarrow \) III  \( \Rightarrow \)   I

Solution: Assuming I is true, then  \(\begin{align}s = \frac{{a + b + c}}{2}\end{align}\)  is rational. Since  \(\Delta \) is also rational.

\[\tan \frac{B}{2} = \frac{\Delta }{{s(s - b)}}\;\;{\text{and}}\;\;\tan \frac{C}{2} = \frac{\Delta }{{s(s - c)}}\]

are also rational. Thus, I \( \Rightarrow \)  II.

Now, assuming II, we see that

\[\sin B = \frac{{2\tan \frac{B}{2}}}{{1 + {{\tan }^2}\frac{B}{2}}}\;\;{\text{and}}\;\,\sin C = \frac{{2\tan \frac{C}{2}}}{{1 + {{\tan }^2}\frac{C}{2}}}\]

are rational. Also,

\[\tan \frac{A}{2} = \cot \left( {\frac{{B + C}}{2}} \right) = \frac{{1 - \tan \frac{B}{2}\tan \frac{C}{2}}}{{\tan \frac{B}{2} + \tan \frac{C}{2}}}\] 

becomes rational, which implies that sin A is rational. Thus, II  \(\Rightarrow \)  III

Finally, assuming III, we have

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

as rational, and further \(\Delta  = \frac{1}{2}bc\;\sin A\) , as rational. Thus, III  \(\Rightarrow \) I.

Example - 53

Prove that  \({s^2} > 4\Delta \)

Solution: Applying the A.M – G. M. inequality on \(s,\;s - a,\;s - b\;{\text{and}}\;s - c,\)

we have

\[\begin{align}&\frac{{s + (s - a) + (s - b) + (s - c)}}{4} > {\left( {s(s - a)(s - b)(s - c)} \right)^{1/4}}\\   \Rightarrow\quad   &\frac{s}{2} > \sqrt \Delta     \quad\Rightarrow\quad {s^2} > 4\Delta \\ \end{align} \]

Note that equality can never hold since s cannot equal the other terms.

Example - 54

Prove that  \(\begin{align}(abcs)\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} = {\Delta ^2}\end{align}\)

Solution: Using the half-angle sine formulae should yield the desired result:

\[\begin{align}&{\text{LHS}} = abcs\sqrt {\frac{{(s - b)(s - c)}}{{bc}}} \sqrt {\frac{{(s - c)(s - a)}}{{ca}}} \sqrt {\frac{{(s - a)(s - b)}}{{ab}}}  \\  \,\,\,\,\,\,\,\,\,\,\,\, &\qquad= \frac{{abc\;s(s - a)(s - b)(s - c)}}{{abc}} \\  \,\,\,\,\,\,\,\,\,\,\,\, &\qquad= {\Delta ^2} \\ \end{align} \]

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Download SOLVED Practice Questions of Examples on Semiperimeter and Half Angle Formulae for FREE
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