### 3D angles

Question Sample Titled '3D angles'

 Figure (a) shows a piece of triangular paper card ${A}{B}{C}$ with ${A}{B}={36}$ $\text{cm}$ , ${B}{C}={15}$ $\text{cm}$ and ${A}{C}={39}$ $\text{cm}$ . Let ${E}$ be a point lying on ${A}{C}$ such that $\angle{B}{E}{C}={76}^{\circ}$.

${A}$${B}$${C}$${E}$Figure (a)

 Find (i) $\angle{B}{C}{E}$ , (ii) ${C}{E}$ .

  (3 marks) (b) Chloe folds the triangular paper card described in (a) along ${B}{E}$ such that ${A}{B}$ and ${B}{C}$ lie on the horizontal ground as shown in Figure (b). It is given that $\angle{A}{E}{C}={159}^{\circ}$ .

Figure (b)${A}$${B}$${C}$${E}$

 (i) Find the distance between ${A}$ and ${C}$ on the horizontal ground. (ii) Let ${F}$ be a point lying on ${B}{C}$ such that ${E}{F}$ is perpendicular to ${B}{C}$ . Chloe claims that the angle between the face ${B}{C}{E}$ and the horizontal ground is $\angle{A}{F}{E}$ . Do you agree? Explain your answer.  (5 marks)

 (ai) Note that $\angle{A}{B}{C}$ $={90}^{\circ}$. ${\tan}\angle{B}{C}{E}$ $=\dfrac{{36}}{{15}}$ $\angle{B}{C}{E}$ $={{\tan}^{{-{1}}}{\left(\tfrac{{36}}{{15}}\right)}}$  $\approx{67.4}^{\circ}$ 1A  (aii) By sine formula, $\dfrac{{{C}{E}}}{{{\sin}\angle{E}{B}{C}}}$ $=\dfrac{{{B}{C}}}{{{\sin}\angle{B}{E}{C}}}$ 1M $\dfrac{{{C}{E}}}{{\sin{{\left({180}^{\circ}-{76}^{\circ}-{{\tan}^{{-{1}}}{\left(\tfrac{{36}}{{15}}\right)}}\right)}}}}$ $\approx\dfrac{{15}}{{{\sin{{76}}}^{\circ}}}$ ${C}{E}$ $=\dfrac{{{15}{\sin{{\left({104}^{\circ}-{{\tan}^{{-{1}}}{\left(\tfrac{{36}}{{15}}\right)}}\right)}}}}}{{{\sin{{76}}}^{\circ}}}$ ${C}{E}$ $\approx{9.22}$ $\text{cm}$ 1A

 (bi) By cosine formula, ${A}{C}^{{2}}$ $={A}{E}^{{2}}+{C}{E}^{{2}}-{2}{\left({A}{E}\right)}{\left({C}{E}\right)}{\cos}\angle{A}{E}{C}$ 1M ${A}{C}$ $\approx{38.5}$ $\text{cm}$ 1A  (bii) ${C}{F}$ $={C}{E}{\cos}\angle{B}{C}{E}$ 1M  $\approx{3.55}$ $\text{cm}$  By cosine formula, ${A}{B}^{{2}}$ $={B}{C}^{{2}}+{A}{C}^{{2}}-{2}{\left({A}{C}\right)}{\left({B}{C}\right)}{\cos}\angle{A}{C}{B}$ $\angle{A}{C}{B}$ $\approx{69.0}^{\circ}$  By cosine formula, ${A}{F}^{{2}}$ $={C}{F}^{{2}}+{A}{C}^{{2}}-{2}{\left({C}{F}\right)}{\left({A}{C}\right)}{\cos}\angle{A}{C}{B}$ ${A}{F}$ $\approx{37.4}$ $\text{cm}$  By sine formula, $\dfrac{{{A}{C}}}{{{\sin}\angle{A}{F}{C}}}$ $=\dfrac{{{A}{F}}}{{{\sin}\angle{A}{C}{F}}}$ $\angle{A}{F}{C}$ $\approx{74.1}^{\circ}$ or $\angle{A}{F}{C}\approx{106}^{\circ}$ So, $\angle{A}{F}{C}$ is not a right angle. 1M Hence, $\angle{A}{F}{E}$ is not the angle between the face ${B}{C}{E}$ and the horizontal ground. Thus, she is incorrect. 1A

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