The distance between vertex and orthocentre of a triangle

In $ \displaystyle \Delta \ ABC,\ \angle A={{60}^{\circ }},\ BC=5$ ,then find the distance of the vertex A from the orthocentre of  $ \displaystyle \Delta \ ABC\ $ .

Solution

$ \displaystyle BF=5\sin \theta \ \ \ ,\ \ CF=5\cos \theta $ 

$ \displaystyle GF=5\sin \theta \tan {{30}^{\circ }}$

$ \displaystyle \ \ \ \ \ \ =\frac{5}{{\sqrt{3}}}\sin \theta $ 

$ \displaystyle \text{By}\ \text{the}\ \text{law}\ \text{of}\ \text{sines}\ ,$ 

$ \displaystyle \frac{{AG}}{{\sin {{{30}}^{\circ }}}}=\frac{{CG}}{{\sin \ ({{{60}}^{\circ }}-\theta )}}$ 

$ \displaystyle \ \ \ 2AG=\frac{{CF-GF}}{{\sin {{{60}}^{\circ }}\cos \theta -\cos {{{60}}^{\circ }}\sin \theta }}$

$ \displaystyle \ \ \ 2AG=\frac{{5\cos \theta -\frac{5}{{\sqrt{3}}}\sin \theta }}{{\frac{{\sqrt{3}}}{2}\cos \theta -\frac{1}{2}\sin \theta }}$

$ \displaystyle \ \ \ \ \ AG=\frac{{\frac{5}{{\sqrt{3}}}(\sqrt{3}\cos \theta -\sin \theta )}}{{\sqrt{3}\cos \theta -\sin \theta }}$

$ \displaystyle \ \ \therefore \ AG=\frac{5}{{\sqrt{3}}}$

Change of base

Suppose that  $ \displaystyle a,b\ \text{and}\ c$ are real numbers greater than 1 .Find the value of $ \displaystyle \frac{1}{{1+{{{\log }}_{{{{a}^{2}}b}}}\left( {\frac{c}{a}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{b}^{2}}c}}}\left( {\frac{a}{b}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{c}^{2}}a}}}\left( {\frac{b}{c}} \right)}}$ .

Solution

$ \displaystyle \frac{1}{{1+{{{\log }}_{{{{a}^{2}}b}}}\left( {\frac{c}{a}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{b}^{2}}c}}}\left( {\frac{a}{b}} \right)}}+\frac{1}{{1+{{{\log }}_{{{{c}^{2}}a}}}\left( {\frac{b}{c}} \right)}}$ 

$ \displaystyle =\frac{1}{{{{{\log }}_{{{{a}^{2}}b}}}{{a}^{2}}b+{{{\log }}_{{{{a}^{2}}b}}}\left( {\frac{c}{a}} \right)}}+\frac{1}{{{{{\log }}_{{{{b}^{2}}c}}}{{b}^{2}}c+{{{\log }}_{{{{b}^{2}}c}}}\left( {\frac{a}{b}} \right)}}+\frac{1}{{{{{\log }}_{{{{c}^{2}}a}}}{{c}^{2}}a+{{{\log }}_{{{{c}^{2}}a}}}\left( {\frac{b}{c}} \right)}}$ 

$ \displaystyle =\frac{1}{{{{{\log }}_{{{{a}^{2}}b}}}abc}}+\frac{1}{{{{{\log }}_{{{{b}^{2}}c}}}abc}}+\frac{1}{{{{{\log }}_{{{{c}^{2}}a}}}abc}}$ 

$ \displaystyle ={{\log }_{{abc}}}{{a}^{2}}b+{{\log }_{{abc}}}{{b}^{2}}c+{{\log }_{{abc}}}{{c}^{2}}a$

$ \displaystyle ={{\log }_{{abc}}}{{(abc)}^{3}}$ 

$ \displaystyle =3$

Solving problem with Double Angle Formula

$ \displaystyle \text{Prove}\ \text{that}\ \ \ \frac{1}{2}\sqrt{{4{{{\sin }}^{2}}{{{36}}^{\circ }}-1}}=\cos {{72}^{\circ }}.$

$ \displaystyle \text{Solution}$

$ \displaystyle \ \ \ \cos {{72}^{\circ }}=2{{\cos }^{2}}{{36}^{\circ }}-1\ \ \ ........(1)$

$ \displaystyle \ \ \ \cos {{144}^{\circ }}=2{{\cos }^{2}}{{72}^{\circ }}-1$ 

$ \displaystyle \ \ \ -\cos {{36}^{\circ }}=2{{\cos }^{2}}{{72}^{\circ }}-1$

$ \displaystyle \ \ \ \ \ \cos {{36}^{\circ }}=1-2{{\cos }^{2}}{{72}^{\circ }}\ \ .......(2)$

$ \displaystyle (1)+(2)\Rightarrow \ \ \ \cos {{72}^{\circ }}+\cos {{36}^{\circ }}=2\ (\ {{\cos }^{2}}{{36}^{\circ }}-{{\cos }^{2}}{{72}^{\circ }})$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cos {{36}^{\circ }}-\cos {{72}^{\circ }}=\frac{1}{2}$

$ \displaystyle (2)-(1)\Rightarrow \ \ \ \cos {{36}^{\circ }}-\cos {{72}^{\circ }}=2-2\ (\ {{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{72}^{\circ }})$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}=2-2\ (\ {{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{72}^{\circ }})$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ (\ {{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{72}^{\circ }})=\frac{3}{2}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{72}^{\circ }}=\frac{3}{4}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1-{{\sin }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{72}^{\circ }}=\frac{3}{4}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{\sin }^{2}}{{36}^{\circ }}-{{\cos }^{2}}{{72}^{\circ }}=\frac{1}{4}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4{{\sin }^{2}}{{36}^{\circ }}-4{{\cos }^{2}}{{72}^{\circ }}=1$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4{{\sin }^{2}}{{36}^{\circ }}-1=4{{\cos }^{2}}{{72}^{\circ }}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{{4{{{\sin }}^{2}}{{{36}}^{\circ }}-1}}=2\cos {{72}^{\circ }}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ \ \frac{1}{2}\sqrt{{4{{{\sin }}^{2}}{{{36}}^{\circ }}-1}}=\cos {{72}^{\circ }}$

Positive real numbers and their equations

Given : $ \displaystyle a,b$ are positive real numbers such that $ \displaystyle a\sqrt{a}+b\sqrt{b}=183$ and $ \displaystyle a\sqrt{b}+b\sqrt{a}=182$ .
Find : $ \displaystyle \frac{9}{5}\ (\ a+b\ )$

Solution

$ \displaystyle a\sqrt{a}+b\sqrt{b}=183$

$ \displaystyle {{(\ \sqrt{a}\ )}^{3}}+{{(\ \sqrt{b}\ )}^{3}}=183$

$ \displaystyle {{(\ \sqrt{a}+\sqrt{b}\ )}^{3}}-3\sqrt{{ab}}(\ \sqrt{a}+\sqrt{b}\ )=183\ ........(1)$

$ \displaystyle a\sqrt{b}+b\sqrt{a}=182$ 

$ \displaystyle \sqrt{{ab}}\ (\ \sqrt{a}+\sqrt{b}\ )=182\ \ .............(2)$ 

$ \displaystyle \text{From}\ (1)\ \text{and}\ (2),$

$ \displaystyle {{(\ \sqrt{a}+\sqrt{b}\ )}^{3}}-546=183$

$ \displaystyle \ \ \ \ \ \ \ \ {{(\ \sqrt{a}+\sqrt{b}\ )}^{3}}=729$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{a}+\sqrt{b}=9\ \ \ .............(3)$

$ \displaystyle \text{From}\ (2)\ \text{and}\ (3),$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \sqrt{{ab}}=\frac{{182}}{9}$

$ \displaystyle {{(\ \sqrt{a}+\sqrt{b}\ )}^{2}}=81\ \ \ \ \left[ {By\ (3)} \right]$

$ \displaystyle a+b+2\sqrt{{ab}}=81$

$ \displaystyle \ \ a+b+\frac{{364}}{9}=81$

$ \displaystyle \ a+b=\frac{{365}}{9}=73\times \frac{5}{9}$

$ \displaystyle \therefore \ \ \frac{9}{5}\ (\ a+b\ )=73$

 

Solving problem with basic trigonometry formula and double angle formula

$ \displaystyle \text{Prove}\ \text{that}\ \operatorname{cosec}\frac{{{{{180}}^{\circ }}}}{7}=\ \operatorname{cosec}\frac{{{{{360}}^{\circ }}}}{7}+\ \operatorname{cosec}\frac{{{{{540}}^{\circ }}}}{7}.$


Solution

$ \displaystyle \ \operatorname{cosec}\frac{{{{{360}}^{\circ }}}}{7}+\ \operatorname{cosec}\frac{{{{{540}}^{\circ }}}}{7}=\frac{1}{{\sin \frac{{2\pi }}{7}}}+\frac{1}{{\sin \frac{{3\pi }}{7}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sin \frac{{3\pi }}{7}+\sin \frac{{2\pi }}{7}}}{{\sin \frac{{2\pi }}{7}\sin \frac{{3\pi }}{7}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2\sin \frac{{5\pi }}{{14}}\cos \frac{\pi }{{14}}}}{{2\sin \frac{\pi }{7}\cos \frac{\pi }{7}\cos \frac{\pi }{{14}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\cos \frac{\pi }{7}}}{{\sin \frac{\pi }{7}\cos \frac{\pi }{7}}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \operatorname{cosec}\frac{\pi }{7}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \operatorname{cosec}\frac{{{{{180}}^{\circ }}}}{7}$

Solving problem with Sines Law and Compound Angle Formula

X  is any point on the side BC of a triangle ABC . If the $ \displaystyle \angle AXC=\theta \ and\ \frac{{BX}}{{XC}}=\frac{m}{n}$ , show that $ \displaystyle (m+n)\cos \theta =n\cot B-m\cot C$ .

Solution

$ \displaystyle \text{Proof}:\ \ \text{By}\ \text{the}\ \text{law}\ \text{of}\ \text{sines}\ ,$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \frac{{BX}}{{\sin \beta }}=\frac{{AB}}{{\sin ({{{180}}^{\circ }}-\theta )}}=\frac{{AB}}{{\sin \theta }}\ \ .....(1)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \frac{{XC}}{{\sin \alpha }}=\frac{{AC}}{{\sin \theta }}\ \ \ ..........(2)$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \frac{{AB}}{{\sin C}}=\frac{{AC}}{{\sin B}}\ \ \ ..........(3)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ (1)\div (2)\Rightarrow \ \ \ \frac{{BX\sin \alpha }}{{XC\sin \beta }}=\frac{{AB}}{{AC}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{m\sin \alpha }}{{n\sin \beta }}=\frac{{\sin C}}{{\sin B}}\ \ \ \ [\text{By}\ (3)]$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{m\sin ({{{180}}^{\circ }}-(\theta +C))}}{{n\sin (\theta -B)}}=\frac{{\sin C}}{{\sin B}}\ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{m\sin (\theta +C)}}{{n\sin (\theta -B)}}=\frac{{\sin C}}{{\sin B}}\ $ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \frac{{m(\sin \theta \cos C+\cos \theta \sin C)}}{{\sin C}}=\frac{{n(\sin \theta \cos B-\cos \theta \sin B)}}{{\sin B}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m\sin \theta \cot C+m\cos \theta =n\sin \theta \cot B-n\cos \theta $ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (m+n)\cos \theta \ =\sin \theta \ (\ n\cot B-m\cot C\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ \ \ \ (m+n)\cot \theta \ =\ n\cot B-m\cot C$

Solving problem with Sines Law and Cosine Law

In triangle ABC, the lengths of the three sides of the triangle are $ \displaystyle a\ cm,b\ cm\ and\ c\ cm$ . It is given that $ \displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{{{c}^{2}}}}=2016$. Find the value of $ \displaystyle \frac{{\cot C}}{{\cot A+\cot B}}$ .

Solution

$ \displaystyle \frac{{{{a}^{2}}+{{b}^{2}}}}{{{{c}^{2}}}}=2016$ 

$ \displaystyle {{a}^{2}}+{{b}^{2}}=2016\ {{c}^{2}}$ 

$ \displaystyle {{a}^{2}}+{{b}^{2}}-{{c}^{2}}=2015\ {{c}^{2}}$ 

$ \displaystyle 2ab\cos C=2015\ {{c}^{2}}\ \ (\because \ \ \cos C=\frac{{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}}{{2ab}})$

$ \displaystyle \ \ \ \ \ \ \cos C=\frac{{2015\ {{c}^{2}}}}{{2ab}}$ 

$ \displaystyle \frac{{\cot C}}{{\cot A+\cot B}}=\frac{{\frac{{\cos C}}{{\sin C}}}}{{\frac{{\cos A}}{{\sin A}}+\frac{{\cos B}}{{\sin B}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\frac{{\cos C}}{{\sin C}}}}{{\frac{{\sin B\cos A+\cos B\sin A}}{{\sin A\sin B}}}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\cos C}}{{\sin C}}\times \frac{{\sin A\sin B}}{{\sin (B+A)}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\cos C}}{{\sin C}}\times \frac{{\sin A\sin B}}{{\sin ({{{180}}^{\circ }}-C)}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\cos C}}{{\sin C}}\times \frac{{\sin A\sin B}}{{\sin C}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos C\times \frac{{\sin A}}{{\sin C}}\times \frac{{\sin B}}{{\sin C}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2015\ {{c}^{2}}}}{{2ab}}\times \frac{a}{c}\times \frac{b}{c}\ \ (\because \ \frac{a}{{\sin A}}=\frac{b}{{\sin B}}=\frac{c}{{\sin C}})$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2015}}{2}$