### 3D angles in pyramid

Question Sample Titled '3D angles in pyramid'

Figure (a) shows a right pyramid ${T}{A}{B}{C}{D}$ with a square base, where $\angle{T}{A}{B}={75}^{\circ}$ . The length of a side of the base is ${23}$ $\text{cm}$ . Let ${H}$ and ${K}$ be the points lying on ${T}{A}$ and ${T}{B}$ respectively such that ${H}{K}$ is parallel to ${C}{D}$ and $\angle{H}{D}{A}={57}^{\circ}$ . A geometric model is made by cutting off the pyramid ${T}{H}{K}{C}{D}$ from ${T}{A}{B}{C}{D}$ as shown in Figure (b).

${T}$${A}$${C}$${D}$${B}$${T}$${A}$${C}$${D}$${B}$${H}$${K}$Figure (a)Figure (b)

 (a) Find the length of ${A}{H}$ . (2 marks) (b) Let $\theta$ be the angle between the plane ${H}{D}{C}{K}$ and the base ${A}{B}{C}{D}$. (i) Find $\theta$ . (ii) Let $\phi$ be the angle between ${H}{D}$ and the base ${A}{B}{C}{D}$ . Which one of $\theta$ and $\phi$ is greater? Explain your answer.  (6 marks)

 (a) By sine formula, $\dfrac{{{A}{H}}}{{{\sin}\angle{H}{D}{A}}}$ $=\dfrac{{{A}{D}}}{{{\sin}\angle{A}{H}{D}}}$ 1M $\dfrac{{{A}{H}}}{{{\sin{{57}}}^{\circ}}}$ $=\dfrac{{{23}}}{{{\sin{{\left({180}^{\circ}-{57}^{\circ}-{75}^{\circ}\right)}}}}}$ ${A}{H}$ $=\dfrac{{{23}{\sin{{57}}}^{\circ}}}{{{\sin{{48}}}^{\circ}}}$ $\therefore$Thus, the length of ${A}{H}$ is ${26.0}$ $\text{cm}$ . 1A  (bi) Let ${X}$ be the foot of the perpendicular from ${H}$ at ${A}{B}$ . ${H}{X}$ $={A}{H}{\sin}\angle{H}{A}{B}$ 1M  $=\dfrac{{{23}{{\sin{{57}}}^{\circ}{\sin{{75}}}^{\circ}}}}{{{\sin{{48}}}^{\circ}}}$ ${A}{X}$ $={A}{H}{\cos}\angle{H}{A}{B}$  $=\dfrac{{{23}{{\sin{{57}}}^{\circ}{\cos{{75}}}^{\circ}}}}{{{\sin{{48}}}^{\circ}}}$ By sine formula, $\dfrac{{{H}{D}}}{{{\sin}\angle{H}{A}{D}}}$ $=\dfrac{{{A}{D}}}{{{\sin}\angle{A}{H}{D}}}$ ${H}{D}$ $=\dfrac{{{23}{\sin{{75}}}^{\circ}}}{{{\sin{{48}}}^{\circ}}}$  Let ${Y}$ be the foot of the perpendicular from ${H}$ to ${C}{D}$ . ${H}{Y}^{{2}}$ $={H}{D}^{{2}}-{A}{X}^{{2}}$ ${H}{Y}$ $=\sqrt{{{\left(\dfrac{{{23}{\sin{{75}}}^{\circ}}}{{{\sin{{48}}}^{\circ}}}\right)}^{{2}}-{\left(\dfrac{{{23}{{\sin{{57}}}^{\circ}{\cos{{75}}}^{\circ}}}}{{{\sin{{48}}}^{\circ}}}\right)}^{{2}}}}$  Note that $\theta$ $=\angle{X}{Y}{H}$ . 1M By cosine formula, ${\cos{\theta}}$ $=\dfrac{{{H}{Y}^{{2}}+{X}{Y}^{{2}}-{H}{X}^{{2}}}}{{{2}{\left({H}{Y}\right)}{\left({X}{Y}\right)}}}$ 1M $\theta$ $\approx{56.0}^{\circ}$ 1A  (bii) Let ${Z}$ be the projection of ${H}$ on the base ${A}{B}{C}{D}$. Then, $\phi$ $=\angle{H}{D}{Z}$. 1M Note that ${H}{D}$ $>{H}{Y}$ . ${\sin{\theta}}$ $=\dfrac{{{H}{Z}}}{{{H}{Y}}}$  $>\dfrac{{{H}{Z}}}{{{H}{D}}}$  $={\sin}\angle{H}{D}{Z}$  $=\phi$ Since $\theta$ and $\phi$ are acute angles, $\theta$ is greater than $\phi$. 1A

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