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Question Number 19243 by Tinkutara last updated on 07/Aug/17

Let a, b, c be the sides opposite the  angles A, B and C respectively of a  ΔABC. Find the value of k such that  (a) a + b = kc  (b) cot (A/2) + cot (B/2) = k cot (C/2).

$$\mathrm{Let}\:{a},\:{b},\:{c}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{opposite}\:\mathrm{the} \\ $$$$\mathrm{angles}\:\mathrm{A},\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{respectively}\:\mathrm{of}\:\mathrm{a} \\ $$$$\Delta\mathrm{ABC}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left(\mathrm{a}\right)\:{a}\:+\:{b}\:=\:{kc} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\:+\:\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:=\:{k}\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}}. \\ $$

Commented by Tinkutara last updated on 08/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Commented by ajfour last updated on 07/Aug/17

Commented by ajfour last updated on 08/Aug/17

a+b=kc , from figure this means or  ⇒   (x+y)+(x+z)=k(y+z)  ...(i)    rcot (A/2)+rcot (B/2)=krcot (C/2)  ⇒   z+y = kx   ...(ii)  using (ii) in (i):          2x+kx=k(kx)       ⇒    k^2 −k−2=0               (k−2)(k+1)=0      k=−1, k=2  For the case here ,  k=2 .

$$\mathrm{a}+\mathrm{b}=\mathrm{kc}\:,\:\mathrm{from}\:\mathrm{figure}\:\mathrm{this}\:\mathrm{means}\:\mathrm{or} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{x}+\mathrm{y}\right)+\left(\mathrm{x}+\mathrm{z}\right)=\mathrm{k}\left(\mathrm{y}+\mathrm{z}\right)\:\:...\left(\mathrm{i}\right) \\ $$$$\:\:\mathrm{rcot}\:\frac{\mathrm{A}}{\mathrm{2}}+\mathrm{rcot}\:\frac{\mathrm{B}}{\mathrm{2}}=\mathrm{krcot}\:\frac{\mathrm{C}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\mathrm{z}+\mathrm{y}\:=\:\mathrm{kx}\:\:\:...\left(\mathrm{ii}\right) \\ $$$$\mathrm{using}\:\left(\mathrm{ii}\right)\:\mathrm{in}\:\left(\mathrm{i}\right): \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2x}+\mathrm{kx}=\mathrm{k}\left(\mathrm{kx}\right) \\ $$$$\:\:\:\:\:\Rightarrow\:\:\:\:\mathrm{k}^{\mathrm{2}} −\mathrm{k}−\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{k}−\mathrm{2}\right)\left(\mathrm{k}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{k}=−\mathrm{1},\:\mathrm{k}=\mathrm{2} \\ $$$$\mathrm{For}\:\mathrm{the}\:\mathrm{case}\:\mathrm{here}\:,\:\:\boldsymbol{\mathrm{k}}=\mathrm{2}\:. \\ $$

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