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Question Number 208385 by efronzo1 last updated on 14/Jun/24

Commented by efronzo1 last updated on 15/Jun/24

Commented by efronzo1 last updated on 15/Jun/24

i′m stuck this step

$$\mathrm{i}'\mathrm{m}\:\mathrm{stuck}\:\mathrm{this}\:\mathrm{step} \\ $$

Commented by mr W last updated on 15/Jun/24

⇒y=((2+x)/3) ⇒z=((4x)/3) cos (α+β)=cos α cos β−(√((1−cos^2 α)(1−cos^2 β))) z=xy−(√((1−x^2 )(1−y^2 ))) (xy−z)^2 =(1−x^2 )(1−y^2 ) x^2 y^2 +z^2 −2xyz=1−x^2 −y^2 +x^2 y^2 z^2 −2xyz=1−x^2 −y^2 ((16x^2 )/9)−((8x^2 (2+x))/9)=1−x^2 −(((2+x)^2 )/9) 8x^3 −10x^2 −4x+5=0 ⇒x=±((√2)/2), (5/4) only x=−((√2)/2) is suitable cos α=x=−((√2)/2) ⇒α=135°

$$\Rightarrow{y}=\frac{\mathrm{2}+{x}}{\mathrm{3}} \\ $$$$\Rightarrow{z}=\frac{\mathrm{4}{x}}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta−\sqrt{\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\alpha\right)\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\beta\right)} \\ $$$${z}={xy}−\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right)} \\ $$$$\left({xy}−{z}\right)^{\mathrm{2}} =\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right) \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{xyz}=\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{xyz}=\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$$\frac{\mathrm{16}{x}^{\mathrm{2}} }{\mathrm{9}}−\frac{\mathrm{8}{x}^{\mathrm{2}} \left(\mathrm{2}+{x}\right)}{\mathrm{9}}=\mathrm{1}−{x}^{\mathrm{2}} −\frac{\left(\mathrm{2}+{x}\right)^{\mathrm{2}} }{\mathrm{9}} \\ $$$$\mathrm{8}{x}^{\mathrm{3}} −\mathrm{10}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow{x}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${only}\:{x}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{is}\:{suitable} \\ $$$$\mathrm{cos}\:\alpha={x}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\alpha=\mathrm{135}° \\ $$

Answered by mr W last updated on 14/Jun/24

s^2 =((p^2 +r^2 +(√(4(p^2 +r^2 −q^2 )q^2 −(r^2 −p^2 )^2 )))/2) =((1^2 +3^2 +(√(4(1^2 +3^2 −2^2 )2^2 −(3^2 −1^2 )^2 )))/2) =5+2(√2) cos ∠APB=((1^2 +2^2 −(5+2(√2)))/(2×1×2))=−((√2)/2) ⇒∠APB=135°

$${s}^{\mathrm{2}} =\frac{{p}^{\mathrm{2}} +{r}^{\mathrm{2}} +\sqrt{\mathrm{4}\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} −{q}^{\mathrm{2}} \right){q}^{\mathrm{2}} −\left({r}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\:\:=\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\sqrt{\mathrm{4}\left(\mathrm{1}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} \right)\mathrm{2}^{\mathrm{2}} −\left(\mathrm{3}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\:\:=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{cos}\:\angle{APB}=\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{2}×\mathrm{1}×\mathrm{2}}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow\angle{APB}=\mathrm{135}° \\ $$

Commented by efronzo1 last updated on 14/Jun/24

what

$$\mathrm{what}\:\:\: \\ $$

Commented by mr W last updated on 15/Jun/24

Commented by mr W last updated on 15/Jun/24

cos α=((s^2 +q^2 −p^2 )/(2qs)) sin α=cos β=((s^2 +q^2 −r^2 )/(2qs)) (((s^2 +q^2 −p^2 )/(2qs)))^2 +(((s^2 +q^2 −r^2 )/(2qs)))^2 =1 (s^2 +q^2 −p^2 )^2 +(s^2 +q^2 −r^2 )^2 =4q^2 s^2 2s^4 −2(p^2 +r^2 )s^2 +p^4 +r^4 +2q^4 −2(p^2 +r^2 )q^2 =0 ⇒s^2 =((p^2 +r^2 ±(√(4(p^2 +r^2 −q^2 )q^2 −(r^2 −p^2 )^2 )))/2) + for point P inside the square − for point P outside the square

$$\mathrm{cos}\:\alpha=\frac{{s}^{\mathrm{2}} +{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{2}{qs}} \\ $$$$\mathrm{sin}\:\alpha=\mathrm{cos}\:\beta=\frac{{s}^{\mathrm{2}} +{q}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{qs}} \\ $$$$\left(\frac{{s}^{\mathrm{2}} +{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{2}{qs}}\right)^{\mathrm{2}} +\left(\frac{{s}^{\mathrm{2}} +{q}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{qs}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left({s}^{\mathrm{2}} +{q}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({s}^{\mathrm{2}} +{q}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4}{q}^{\mathrm{2}} {s}^{\mathrm{2}} \\ $$$$\mathrm{2}{s}^{\mathrm{4}} −\mathrm{2}\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){s}^{\mathrm{2}} +{p}^{\mathrm{4}} +{r}^{\mathrm{4}} +\mathrm{2}{q}^{\mathrm{4}} −\mathrm{2}\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\frac{{p}^{\mathrm{2}} +{r}^{\mathrm{2}} \pm\sqrt{\mathrm{4}\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} −{q}^{\mathrm{2}} \right){q}^{\mathrm{2}} −\left({r}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$+\:{for}\:{point}\:{P}\:{inside}\:{the}\:{square} \\ $$$$−\:{for}\:{point}\:{P}\:{outside}\:{the}\:{square} \\ $$

Commented by Tawa11 last updated on 21/Jun/24

Weldone sir.

$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$

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