Question-50755

Question Number 50755 by ajfour last updated on 19/Dec/18

Commented by ajfour last updated on 19/Dec/18

Find the angle between the two triangular (coloured) planes.

$${Find}\:{the}\:{angle}\:{between}\:{the}\:{two} \\ $$$${triangular}\:\left({coloured}\right)\:{planes}. \\ $$

Answered by mr W last updated on 20/Dec/18

let θ=angle between both coloured triangles Δ_(abc) =((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4) Δ_(pqb) =((√((p+q+b)(−p+q+b)(p−q+b)(p+q−b)))/4) Δ_(pqb) =((bh_b )/2) ⇒h_b =((2Δ_(pqb) )/b)=altitude to base line b ⇒h_b =((√((p+q+b)(−p+q+b)(p−q+b)(p+q−b)))/(2b)) V=((Δ_(abc) h)/3) V=((√(a^2 q^2 (−a^2 −q^2 +b^2 +l^2 +c^2 +p^2 )+b^2 l^2 (a^2 +q^2 −b^2 −l^2 +c^2 +p^2 )+c^2 p^2 (a^2 +q^2 +b^2 +l^2 −c^2 −p^2 )−a^2 b^2 c^2 −a^2 l^2 p^2 −b^2 q^2 p^2 −c^2 q^2 l^2 ))/(12)) ⇒h=((3V)/Δ_(abc) )=altitude to base triangle abc h=h_b sin θ ⇒θ=sin^(−1) (h/h_b )=sin^(−1) ((3bV)/(2Δ_(abc) Δ_(pqb) )) ⇒θ=sin^(−1) ((2b(√(a^2 q^2 (−a^2 −q^2 +b^2 +l^2 +c^2 +p^2 )+b^2 l^2 (a^2 +q^2 −b^2 −l^2 +c^2 +p^2 )+c^2 p^2 (a^2 +q^2 +b^2 +l^2 −c^2 −p^2 )−a^2 b^2 c^2 −a^2 l^2 p^2 −b^2 q^2 p^2 −c^2 q^2 l^2 )))/( (√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)(p+q+b)(−p+q+b)(p−q+b)(p+q−b)))))

$${let}\:\theta={angle}\:{between}\:{both}\:{coloured}\:{triangles} \\ $$$$\Delta_{{abc}} =\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$$$\Delta_{{pqb}} =\frac{\sqrt{\left({p}+{q}+{b}\right)\left(−{p}+{q}+{b}\right)\left({p}−{q}+{b}\right)\left({p}+{q}−{b}\right)}}{\mathrm{4}} \\ $$$$\Delta_{{pqb}} =\frac{{bh}_{{b}} }{\mathrm{2}} \\ $$$$\Rightarrow{h}_{{b}} =\frac{\mathrm{2}\Delta_{{pqb}} }{{b}}={altitude}\:{to}\:{base}\:{line}\:{b} \\ $$$$\Rightarrow{h}_{{b}} =\frac{\sqrt{\left({p}+{q}+{b}\right)\left(−{p}+{q}+{b}\right)\left({p}−{q}+{b}\right)\left({p}+{q}−{b}\right)}}{\mathrm{2}{b}} \\ $$$${V}=\frac{\Delta_{{abc}} {h}}{\mathrm{3}} \\ $$$${V}=\frac{\sqrt{{a}^{\mathrm{2}} {q}^{\mathrm{2}} \left(−{a}^{\mathrm{2}} −{q}^{\mathrm{2}} +{b}^{\mathrm{2}} +{l}^{\mathrm{2}} +{c}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} {l}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{q}^{\mathrm{2}} −{b}^{\mathrm{2}} −{l}^{\mathrm{2}} +{c}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} {p}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{q}^{\mathrm{2}} +{b}^{\mathrm{2}} +{l}^{\mathrm{2}} −{c}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} −{a}^{\mathrm{2}} {l}^{\mathrm{2}} {p}^{\mathrm{2}} −{b}^{\mathrm{2}} {q}^{\mathrm{2}} {p}^{\mathrm{2}} −{c}^{\mathrm{2}} {q}^{\mathrm{2}} {l}^{\mathrm{2}} }}{\mathrm{12}} \\ $$$$\Rightarrow{h}=\frac{\mathrm{3}{V}}{\Delta_{{abc}} }={altitude}\:{to}\:{base}\:{triangle}\:{abc} \\ $$$${h}={h}_{{b}} \:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \frac{{h}}{{h}_{{b}} }=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}{bV}}{\mathrm{2}\Delta_{{abc}} \Delta_{{pqb}} } \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} {q}^{\mathrm{2}} \left(−{a}^{\mathrm{2}} −{q}^{\mathrm{2}} +{b}^{\mathrm{2}} +{l}^{\mathrm{2}} +{c}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} {l}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{q}^{\mathrm{2}} −{b}^{\mathrm{2}} −{l}^{\mathrm{2}} +{c}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} {p}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{q}^{\mathrm{2}} +{b}^{\mathrm{2}} +{l}^{\mathrm{2}} −{c}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} −{a}^{\mathrm{2}} {l}^{\mathrm{2}} {p}^{\mathrm{2}} −{b}^{\mathrm{2}} {q}^{\mathrm{2}} {p}^{\mathrm{2}} −{c}^{\mathrm{2}} {q}^{\mathrm{2}} {l}^{\mathrm{2}} }}{\:\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)\left({p}+{q}+{b}\right)\left(−{p}+{q}+{b}\right)\left({p}−{q}+{b}\right)\left({p}+{q}−{b}\right)}} \\ $$

Commented by mr W last updated on 20/Dec/18

Commented by ajfour last updated on 20/Dec/18

$${Great}\:{Sir}!\:{Thanks}. \\ $$

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