Find the side and angle in a slant tetrahedron and determine the angle between a side and a plane

Question Sample Titled 'Find the side and angle in a slant tetrahedron and determine the angle between a side and a plane'

The figure below shows a geometric model ${A}{B}{C}{D}$ in the form of tetrahedron. It is given that $\angle{B}{A}{D}={80}^{\circ}$ , $\angle{C}{B}{D}={41}^{\circ}$ , ${A}{B}={148}$ $\text{cm}$ , ${A}{C}={48}$ $\text{cm}$ , ${B}{C}={140}$ $\text{cm}$ and ${B}{D}={283}$ $\text{cm}$ .

${A}$${B}$${C}$${D}$

 (a) Find $\angle{A}{B}{D}$ and ${C}{D}$ . (4 marks) (b) A craftsman claims that the angle between ${A}{B}$ and the face ${B}{C}{D}$ is $\angle{A}{B}{C}$ . Do you agree? Explain your answer. (2 marks)

 (a) By sine formula, $\dfrac{{{\sin}\angle{B}{A}{D}}}{{{B}{D}}}$ $=\dfrac{{{\sin}\angle{A}{D}{B}}}{{{A}{B}}}$ 1M $\dfrac{{\sin{{80}}}^{\circ}}{{283}}$ $=\dfrac{{{\sin}\angle{A}{D}{B}}}{{148}}$ $\angle{A}{D}{B}$ $\approx{30.99900143}^{\circ}$  $\angle{A}{B}{D}$ $\approx{180}^{\circ}-{80}^{\circ}-{30.99900143}^{\circ}$ $\angle{A}{B}{D}$ $\approx{69.00099857}^{\circ}$ $\angle{A}{B}{D}$ $\approx{69.0}^{\circ}$ 1A  By cosine formula, ${C}{D}^{{2}}$ $={B}{C}^{{2}}+{B}{D}^{{2}}-{2}{\left({B}{C}\right)}{\left({B}{D}\right)}{\left({\cos}\angle{C}{B}{D}\right)}$ 1M ${C}{D}$ $=\sqrt{{{140}^{{2}}+{283}^{{2}}-{2}{\left({140}\right)}{\left({283}\right)}{\left({\cos{{41}}}^{\circ}\right)}}}$ ${C}{D}$ $\approx{199.7143281}$ $\text{cm}$ ${C}{D}$ $\approx{200}$ $\text{cm}$ 1A  (b) By cosine formula, ${A}{D}^{{2}}$ $={A}{B}^{{2}}+{B}{D}^{{2}}-{2}{\left({A}{B}\right)}{\left({B}{D}\right)}{\left({\cos}\angle{A}{B}{D}\right)}$ ${A}{D}$ $\approx\sqrt{{{148}^{{2}}+{283}^{{2}}-{2}{\left({148}\right)}{\left({283}\right)}{\left({\cos{{69.00099857}}}^{\circ}\right)}}}$ ${A}{D}$ $\approx{268.2808166}$ $\text{cm}$  Consider $\triangle{A}{C}{D}$ , ${A}{C}^{{2}}+{C}{D}^{{2}}$ $={48}^{{2}}+{199.7143281}^{{2}}$ $\approx{42189.81285}$ ${A}{D}^{{2}}$ $\approx{268.2808166}^{{2}}$ $\approx{71974.59658}$ ∵  ${A}{C}^{{2}}+{C}{D}^{{2}}\ne{A}{D}^{{2}}$ ∴  $\angle{A}{C}{D}$ is not a right angle. 1M or any correct proof showing the absence of orthogonality of ${A}{C}$ and ${C}{D}$  ∵  $\angle{A}{C}{D}$ is not a right angle, ${A}{C}$ is not perpendicular to plane ${B}{C}{D}$ . Thus, the claim is disagreed. 1A f.t.  The figure below shows the possible situtation in which ${X}$ and ${C}$ not coincident but both lying on place ${B}{C}{D}$ . Notice that in the figure, ${A}{X}\bot{B}{X}$ and ${A}{X}\bot{X}{D}$ , then the angle between ${A}{B}$ and plane ${B}{C}{D}$ is $\angle{A}{B}{X}$ .

${A}$${B}$${C}$${D}$${X}$

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