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Question Number 38593 by ajfour last updated on 27/Jun/18

Commented by ajfour last updated on 27/Jun/18

Find θ in terms of c .

$${Find}\:\theta\:{in}\:{terms}\:{of}\:\boldsymbol{{c}}\:. \\ $$

Commented by MrW3 last updated on 27/Jun/18

(c/(tan θ))=1+tan θ  tan^2  θ+tan θ−c=0  tan θ=(((√(1+4c))−1)/2)  ⇒θ=tan^(−1) (((√(1+4c))−1)/2)

$$\frac{{c}}{\mathrm{tan}\:\theta}=\mathrm{1}+\mathrm{tan}\:\theta \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\theta+\mathrm{tan}\:\theta−{c}=\mathrm{0} \\ $$$$\mathrm{tan}\:\theta=\frac{\sqrt{\mathrm{1}+\mathrm{4}{c}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{1}+\mathrm{4}{c}}−\mathrm{1}}{\mathrm{2}} \\ $$

Commented by ajfour last updated on 27/Jun/18

thank you Sir.

$${thank}\:{you}\:{Sir}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

tanθ=(c/(1+b))     b=perpendicular of smal right  angled triangle  tanθ=(b/1)  (c/(1+b))=(b/1)    b^ +b^2 =c  b^2 +b−c=0  b=((−1+(√(1^2 −4×1×(−c))) )/2)=((−1+(√(1+4c)) )/2)  tanθ=(b/1)=(((√(1+4c)) −1)/2)  pls check

$${tan}\theta=\frac{{c}}{\mathrm{1}+{b}}\:\:\:\:\:{b}={perpendicular}\:{of}\:{smal}\:{right} \\ $$$${angled}\:{triangle} \\ $$$${tan}\theta=\frac{{b}}{\mathrm{1}} \\ $$$$\frac{{c}}{\mathrm{1}+{b}}=\frac{{b}}{\mathrm{1}}\:\: \\ $$$${b}^{} +{b}^{\mathrm{2}} ={c} \\ $$$${b}^{\mathrm{2}} +{b}−{c}=\mathrm{0} \\ $$$${b}=\frac{−\mathrm{1}+\sqrt{\mathrm{1}^{\mathrm{2}} −\mathrm{4}×\mathrm{1}×\left(−{c}\right)}\:}{\mathrm{2}}=\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{c}}\:}{\mathrm{2}} \\ $$$${tan}\theta=\frac{{b}}{\mathrm{1}}=\frac{\sqrt{\mathrm{1}+\mathrm{4}{c}}\:−\mathrm{1}}{\mathrm{2}} \\ $$$${pls}\:{check} \\ $$

Commented by ajfour last updated on 27/Jun/18

Thank you Sir, but i think the  next question is bit challenging!

$${Thank}\:{you}\:{Sir},\:{but}\:{i}\:{think}\:{the} \\ $$$${next}\:{question}\:{is}\:{bit}\:{challenging}! \\ $$

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