Properties of triangle and circles connected with them
Examaples :
1. If a = 6, b =5, c = 9 then find A
Solution : from the cosine rule,
\(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\)
=\(\frac{{25 + 81 - 36}}{{2.5.9}}\)=\(\frac{{70}}{{90}} = \frac{7}{9}\)
2. In \(\Delta ABC,\) show that \(\sum {(b + c)\cos A = 2s} \)
Solution :
= (b+c) cosA+(c+a)cosB+(a+b)cosC
\(= (b\cos A + c\cos B) + (c\cos B + b\cos C) + (a\cos C + c\cos A)\)
\(= a + b + c = 2s\)
=\(\therefore \Sigma (b + c)\,\,\cos A = 2s\)
3. If \(tan{A\over 2}={5\over6}\), \(tan{C\over 2}={2\over5}\)then the relatiion among a, b, c
Solution : \(\tan \frac{A}{2}.\tan \frac{C}{2} = \frac{5}{6}.\frac{2}{5} = \frac{1}{3}\)
i.e., \(\sqrt {\frac{{(s - b)(s - c)}}{{s(s - a)}}} .\sqrt {\frac{{(s - a)(s - b)}}{{s(s - c)}}} = \frac{1}{3}\)
i.e., \(\frac{{s - b}}{s} = \frac{1}{3} \Rightarrow 3s - 3b = s \Rightarrow 2s = 3b\)
\( \Rightarrow a + c = 2b \Rightarrow a,b,c\) are in A.p
4. Show that \({a^2}\cot A + {b^2}\cot B + {c^2}\cot C = \frac{{abc}}{{4R}}\)
Solution :
\(\begin{gathered}
= {a^2}\cot A + {b^2}\cot B + {c^2}\cot C\, \hfill \\
= \sum {{a^2}\cot A} \hfill \\
\end{gathered} \)
\(= \Sigma 4{R^2}.{\sin ^2}A.\frac{{\cos A}}{{\sin A}}\)
\(= \Sigma 4{R^2}.\sin A.\cos A\)
\(= \Sigma 2{R^2}(2\sin A.\cos A)\)
\(= \Sigma 2{R^2}(\sin 2A)\)
\(= 2{R^2}(\sin 2A + \sin 2B + \sin 2C)\)
\(= 2{R^2}(4\sin A.\sin B.\sin 2C)\)
\(= \frac{1}{R}(2R\sin A)(2R\sin B)(2R\sin C)\)
\(= \frac{{abc}}{R}\)
\(\therefore \,\,\,\,\,\,\,{a^2}\cot A + {b^2}\cot B + {c^2}\cot c = \frac{{abc}}{R}\)
5.If \(a = (b - c)\sec \theta \) , then prove that \(\tan \theta = \frac{{2\sqrt {bc} }}{{b - c}}.\sin \frac{A}{2}\)
solution:\(a = (b - c)\sec \theta \)
\(\Rightarrow sec\theta = \frac{a}{{b - c}}\)
\(\Rightarrow {\tan ^2}\theta = {\sec ^2}\theta - 1\)
\(= \frac{{{a^2}}}{{{{(b - c)}^2}}} - 1\)
\(= \frac{{{a^2} - {{(b - c)}^2}}}{{{{(b - c)}^2}}}\)
\(= \frac{{(a + b - c)(a - b + c)}}{{{{(b - c)}^2}}}\)
\(= \frac{{(2s - 2c).(2s - 2b)}}{{{{(b - c)}^2}}}\)
\(= \frac{{2(s - c).2(s - b)}}{{{{(b - c)}^2}}}\)
\(= \frac{{4bc}}{{{{(b - c)}^2}}}\frac{{(s - c)(s - b)}}{{bc}}\)
\({\tan ^2}\theta = \frac{{4bc}}{{{{(b - c)}^2}}}.{\sin ^2}\frac{A}{2}\)
\(\tan \theta = \frac{{2\sqrt {bc} }}{{b - c}}.\sin \frac{A}{2}\)
6.Show that \(a.{\cos ^2}\frac{A}{2} + b.{\cos ^2}\frac{B}{2} + c.{\cos ^2}\frac{C}{2} = S + \frac{\vartriangle }{R}\)
solution:\(\Sigma a.{\cos ^2}\frac{A}{2} = \frac{1}{2}.\Sigma a(1 + \cos A)\)
\(= \frac{1}{2}.\Sigma (a + a\cos A)\)
\(= \frac{1}{2}[\Sigma a + \Sigma \,a\,cosA]\)
\(= \frac{1}{2}[(a + b + c)] + \frac{1}{2}.\Sigma (2R\,\sin A.\,cosA)\)
=\(= \frac{1}{2}[(a + b + c)] + \frac{R}{2}\Sigma \sin 2A\)
=\(= \frac{1}{2}(2s) + \frac{R}{2}(\sin 2A + \sin 2B + \sin 2C)\)
\(= s + \frac{R}{2}(4\sin A.\sin B.\sin C)\)
\(= s + \frac{1}{R}(2{R^2}\sin A.\sin B.\sin C)\)
\(= s + \frac{1}{R}.\Delta = s + \frac{\Delta }{R}\)
\(\therefore \,\,\,\,\Sigma a.{\cos ^2}\frac{A}{2} = s + \frac{\Delta }{R}\)
7. Prove \(\Sigma {a^3}\cos (B - C) = 3abc\)
Solution :\(\Sigma {a^3}\cos (B - C)\)
\(= \Sigma {a^2}.[a.\cos (B - C)]\)
\(= \Sigma {a^2}.(2R\sin A).(\cos (B - C)\)
=\(= R.\Sigma {a^2}[2\sin (B + C).\cos (B - C)]\)
\(= R\sum\limits_{}^{} {{a^2}\left( {\sin 2B + \sin 2C} \right)} \)
=\(= R.\Sigma {a^2}(2\sin B.\cos B + 2\sin C.\cos C)]\)
\(= \Sigma [{a^2}(2R\sin B)\cos B + {a^2}(2R\sin C).\cos C]\)
\(= \Sigma ({a^2}b\cos B + {a^2}c\cos C]\)
\(= \Sigma ({a^2}b\cos B) + \Sigma ({a^2}c\cos C)\)
\(= {a^2}b\cos B + {b^2}c\cos C + {c^2}a\cos A\)\(+ {a^2}c\cos C + {b^2}a\cos A + {c^2}b\cos B\)
\(= ({a^2}b\cos B + {b^2}a\cos A) + ({b^2}c\cos C + {c^2}b\cos B) + ({c^2}a\cos A + {a^2}c\cos C)\)
\(= ab(a\cos B + b\cos A) + bc(b\cos C + c\cos B) + ca(c\cos A + a\cos C)\)
\(= abc + bca + cab = 3abc\)
\(\Sigma {a^3}\cos (B - C) = 3abc\)