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Question Number 192001 by ajfour last updated on 05/May/23

Commented by ajfour last updated on 05/May/23

Find x, in terms of a, b, m=tan θ.

$${Find}\:{x},\:{in}\:{terms}\:{of}\:{a},\:{b},\:{m}=\mathrm{tan}\:\theta. \\ $$

Answered by ajfour last updated on 05/May/23

say θ=2φ, x=c cos (φ+sin^(−1) ((a−c)/(a+c))+sin^(−1) ((a−b)/(a+b))+φ) =(((a+b)^2 −(b+c)^2 +(c+a)^2 )/(2(a+b)(c+a))) say if φ=0 (((2(√(ac)))/(a+c)))(((2(√(ab)))/(a+b)))−(((a−c)/(a+c)))(((a−b)/(a+b))) =(((a+b)^2 −(b+c)^2 +(c+a)^2 )/(2(a+b)(c+a))) ⇒ 4a(√(bc))−{a^2 −(b+c)a+bc} =a^2 +a(b+c)−bc ⇒ 2a^2 −4a(√(bc))=0 ⇒ if a≠0 then a=2(√(bc)) ✓

$${say}\:\theta=\mathrm{2}\phi,\:{x}={c} \\ $$$$\mathrm{cos}\:\left(\phi+\mathrm{sin}^{−\mathrm{1}} \frac{{a}−{c}}{{a}+{c}}+\mathrm{sin}^{−\mathrm{1}} \frac{{a}−{b}}{{a}+{b}}+\phi\right) \\ $$$$\:\:\:\:=\frac{\left({a}+{b}\right)^{\mathrm{2}} −\left({b}+{c}\right)^{\mathrm{2}} +\left({c}+{a}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{b}\right)\left({c}+{a}\right)} \\ $$$${say}\:{if}\:\phi=\mathrm{0} \\ $$$$\left(\frac{\mathrm{2}\sqrt{{ac}}}{{a}+{c}}\right)\left(\frac{\mathrm{2}\sqrt{{ab}}}{{a}+{b}}\right)−\left(\frac{{a}−{c}}{{a}+{c}}\right)\left(\frac{{a}−{b}}{{a}+{b}}\right) \\ $$$$=\frac{\left({a}+{b}\right)^{\mathrm{2}} −\left({b}+{c}\right)^{\mathrm{2}} +\left({c}+{a}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{b}\right)\left({c}+{a}\right)} \\ $$$$\Rightarrow\:\:\mathrm{4}{a}\sqrt{{bc}}−\left\{{a}^{\mathrm{2}} −\left({b}+{c}\right){a}+{bc}\right\} \\ $$$$\:\:\:={a}^{\mathrm{2}} +{a}\left({b}+{c}\right)−{bc} \\ $$$$\Rightarrow\:\:\mathrm{2}{a}^{\mathrm{2}} −\mathrm{4}{a}\sqrt{{bc}}=\mathrm{0} \\ $$$$\Rightarrow\:\:{if}\:{a}\neq\mathrm{0}\:\:\:{then} \\ $$$${a}=\mathrm{2}\sqrt{{bc}} \\ $$$$\checkmark \\ $$

Answered by mr W last updated on 05/May/23

Commented by mr W last updated on 05/May/23

cos α=((a−b)/(a+b)) cos β=((a−x)/(a+x)) cos γ=(((a+b)^2 +(a+x)^2 −(b+x)^2 )/(2(a+b)(a+x)))=((a(a+b)+(a−b)x)/(a(a+b)+(a+b)x)) α+β+γ=180+θ γ=180+θ−(α+β) let δ=α−θ=cos^(−1) ((a−b)/(a+b))−θ cos γ=−cos (δ+β)=−cos δ cos β+sin δ sin β ⇒((a(a+b)+(a−b)x)/(a(a+b)+(a+b)x))+(((a−x)/(a+x))) cos δ−sin δ (√(1−(((a−x)/(a+x)))^2 ))=0

$$\mathrm{cos}\:\alpha=\frac{{a}−{b}}{{a}+{b}} \\ $$$$\mathrm{cos}\:\beta=\frac{{a}−{x}}{{a}+{x}} \\ $$$$\mathrm{cos}\:\gamma=\frac{\left({a}+{b}\right)^{\mathrm{2}} +\left({a}+{x}\right)^{\mathrm{2}} −\left({b}+{x}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{b}\right)\left({a}+{x}\right)}=\frac{{a}\left({a}+{b}\right)+\left({a}−{b}\right){x}}{{a}\left({a}+{b}\right)+\left({a}+{b}\right){x}} \\ $$$$\alpha+\beta+\gamma=\mathrm{180}+\theta \\ $$$$\gamma=\mathrm{180}+\theta−\left(\alpha+\beta\right) \\ $$$${let}\:\delta=\alpha−\theta=\mathrm{cos}^{−\mathrm{1}} \frac{{a}−{b}}{{a}+{b}}−\theta \\ $$$$\mathrm{cos}\:\gamma=−\mathrm{cos}\:\left(\delta+\beta\right)=−\mathrm{cos}\:\delta\:\mathrm{cos}\:\beta+\mathrm{sin}\:\delta\:\mathrm{sin}\:\beta \\ $$$$\Rightarrow\frac{{a}\left({a}+{b}\right)+\left({a}−{b}\right){x}}{{a}\left({a}+{b}\right)+\left({a}+{b}\right){x}}+\left(\frac{{a}−{x}}{{a}+{x}}\right)\:\mathrm{cos}\:\delta−\mathrm{sin}\:\delta\:\sqrt{\mathrm{1}−\left(\frac{{a}−{x}}{{a}+{x}}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$

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