Question

Solve this Above Question ​

Answer 1

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✪✪ Hence Answered ✪✪

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Answer 2

Given:

A triangle with vertices A(x₁,x₁tanθ₁), B(x₂,x₂tanθ₂) and C(x₃,x₃tanθ₃) whose circumcenter coincides with origin(0,0) and orthocenter is given by H(a,b)

To Prove:

$\bf{\dfrac{a}{b}=\dfrac{cos\theta_{1}+cos\theta_{2}+cos\theta_{3}}{sin\theta_{1}+sin\theta_{2}+sin\theta_{3}}}$

Things to know before solving

• Coordinates of the Centroid(G) of a triangle having vertices A(x₁,y₁), B(x₂,y₂) and C(x₃,y₃) is given by

$\longrightarrow\bf{G=\bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}$

• Orthocenter(H), Centroid(G) and Circumcenter(O) of triangle are collinear and centroid divides the line joining orthocentere and circumcenter in the ratio 2:1.

i.e.    $\setlength{\unitlength}{0.9 cm}}\begin{picture}(12,4)\thicklines\put(6,6){\line(1,0){5.5}}\put(5.6,5.9){ H }\put(11.7,5.9){ O }\put(9,6){\circle*{0.2}}\put(8.8,6.3){ G }\put(7.6,6.3){ 2 }\put(10.3,6.3){ 1 }\put(11.7,5.9){ O }\end{picture}$

• For a ΔABC having circumcentre O,

⟼ OA = OB = OC

Solution:

Let the circumcenter of given triangle be 'O'.

For the given triangle, Coordinates of G is

$\rightarrow\mathrm{\bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{x_{1}tan\theta_{1}+x_{2}tan\theta_{2}+x_{3}tan\theta_{3}}{3}\bigg)}$

Also,

In the given triangle, circumcenter coincides with origin i.e. coordinates of circumcenter(O) are (0,0)

Now, on applying sectional formula on orthocenter(H), centroid(G) and circumcenter(O) of the given triangle, we get

$\setlength{\unitlength}{0.9 cm}}\begin{picture}(12,4)\thicklines\put(6,6){\line(1,0){5.5}}\put(5.6,5.9){ H }\put(11.7,5.9){ O }\put(5,5.5){ (a\,,\,b) }\put(11.4,5.5){ (0\,,\,0) }\put(9,6){\circle*{0.2}}\put(8.8,6.3){ G }\put(6,5){ (\frac{x_{1}+x_{2}+x_{3}}{3}\,,\,\frac{x_{1}tan\theta_{1}+x_{2}tan\theta_{2}+x_{3}tan\theta_{3}}{3}) }\put(7.6,6.3){ 2 }\put(10.3,6.3){ 1 }\put(11.7,5.9){ O }\end{picture}$

$\sf{\bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{x_{1}tan\theta_{1}+x_{2}tan\theta_{2}+x_{3}tan\theta_{3}}{3}\bigg)=\bigg(\dfrac{2(0)+1(a)}{3},\dfrac{2(0)+1(b)}{3}\bigg)}$

$\sf{\bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{x_{1}tan\theta_{1}+x_{2}tan\theta_{2}+x_{3}tan\theta_{3}}{3}\bigg)=\bigg(\dfrac{a}{3},\dfrac{b}{3}\bigg)}$

On comparing above coordinates, we get

$\mathrm{\dfrac{x_{1}+x_{2}+x_{3}}{3}=\dfrac{a}{3}}$

$\mathrm{x_{1}+x_{2}+x_{3}=a}----------(1)$

$\mathrm{\dfrac{x_{1}tan\theta_{1}+x_{2}tan\theta_{2}+x_{3}tan\theta_{3}}{3}=\dfrac{b}{3}}$

$\mathrm{x_{1}tan\theta_{1}+x_{2}tan\theta_{2}+x_{3}tan\theta_{3}=b}---(2)$

On dividing (1) by (2), we get

$\mathrm{\dfrac{x_{1}+x_{2}+x_{3}}{x_{1}tan\theta_{1}+x_{2}tan\theta_{2}+x_{3}tan\theta_{3}}=\dfrac{a}{b}}--(3)$

Now,

Since, O is the circumcenter of given triangle

∴ OA=OB=OC= a(Let)

$\longrightarrow\mathrm{OA=a}$

So, by using distance formula

$\longrightarrow\mathrm{\sqrt{(x_{1}-0)^{2}+(x_{1}tan\theta_{1})^{2}} =a}$

$\longrightarrow\mathrm{\sqrt{x_{1}^{2}+x_{1}^{2}tan\theta^{2}_{1}} =a}$

$\longrightarrow\mathrm{\sqrt{x_{1}^{2}(1+tan\theta^{2}_{1})} =a}$

$\longrightarrow\mathrm{\sqrt{x_{1}^{2}(sec\theta^{2}_{1})} =a}$

On squaring both the sides, we get

$\longrightarrow\mathrm{x_{1}^{2}sec\theta^{2}_{1} =a^{2}}$

$\longrightarrow\mathrm{x_{1}^{2}=\dfrac{a^{2}}{sec\theta^{2}_{1}}}$

$\longrightarrow\mathrm{x_{1}^{2}=a^{2}cos\theta^{2}_{1}}$

$\longrightarrow\mathrm{x_{1}=\sqrt{a^{2}cos\theta^{2}_{1}}}$

$\longrightarrow\mathrm{x_{1}=acos\theta_{1}}$

Similarly,

$\longrightarrow\mathrm{x_{2}=acos\theta_{2}}$

$\longrightarrow\mathrm{x_{3}=acos\theta_{3}}$

On putting values of x₁, x₂ and x₃ in (3), we get

$\sf{\dfrac{acos\theta_{2}+acos\theta_{2}+acos\theta_{3}}{acos\theta_{1}tan\theta_{1}+acos\theta_{2}tan\theta_{2}+acos\theta_{3}tan\theta_{3}}=\dfrac{a}{b}}$

$\sf{\dfrac{a(cos\theta_{2}+cos\theta_{2}+cos\theta_{3})}{a\cancel{cos\theta_{1}}.\dfrac{sin\theta_{1}}{\cancel{cos\theta_{1}}} +a\cancel{cos\theta_{2}}.\dfrac{sin\theta_{2}}{\cancel{cos\theta_{2}}}+a\cancel{cos\theta_{3}}.\dfrac{sin\theta_{3}}{\cancel{cos\theta_{3}}}}=\dfrac{a}{b}}$

$\sf{\dfrac{a(cos\theta_{2}+cos\theta_{2}+cos\theta_{3})}{asin\theta_{1} +asin\theta_{2}+asin\theta_{3}}=\dfrac{a}{b}}$

$\sf{\dfrac{\cancel{a}(cos\theta_{2}+cos\theta_{2}+cos\theta_{3})}{\cancel{a}(sin\theta_{1} +sin\theta_{2}+sin\theta_{3})}=\dfrac{a}{b}}$

$\sf{\dfrac{a}{b}=\dfrac{cos\theta_{2}+cos\theta_{2}+cos\theta_{3}}{sin\theta_{1} +sin\theta_{2}+sin\theta_{3}}}$

Hence Proved