### Angle in a right triangular prism

Question Sample Titled 'Angle in a right triangular prism'

In the figure, ${A}{B}{C}{D}{E}{F}$ is a right triangular prism. It is given that ${A}{B}={36}$ $\text{cm}$ , ${B}{F}={9}$ $\text{cm}$ and $\angle{B}{C}{F}={20}^{\circ}$ . ${M}$ is a point on ${C}{D}$ such that ${D}{M}={30}$ $\text{cm}$ and ${B}{M}={28}$ $\text{cm}$ . Find the angle between ${E}{M}$ and ${B}{M}$ correct to the nearest ${0.01}^{\circ}$.

${E}$${F}$${A}$${C}$${D}$${B}$${M}$${28}$ $\text{cm}$
A
${63.43}^{\circ}$
B
${62.89}^{\circ}$
C
${66.43}^{\circ}$
D
${67.40}^{\circ}$

 Join ${A}{M}$ and ${B}{E}$ . In $\triangle{B}{C}{F}$, ${\tan}\angle{B}{C}{F}$ $=\dfrac{{{B}{F}}}{{{B}{C}}}$ ${{\tan{{20}}}^{\circ}}$ $=\dfrac{{9}}{{{B}{C}}}$ ${B}{C}$ $=\dfrac{{9}}{{{\tan{{20}}}^{\circ}}}$ $\text{cm}$  In $\triangle{D}{A}{M}$, ${A}{M}^{{2}}$ $={A}{D}^{{2}}+{D}{M}^{{2}}$ Pyth. theorem ${A}{M}$ $=\sqrt{{{\left(\dfrac{{9}}{{{\tan{{20}}}^{\circ}}}\right)}^{{2}}+{30}^{{2}}}}$  $={38.87723248642392}$ $\text{cm}$  In $\triangle{A}{E}{M}$, ${E}{M}^{{2}}$ $={A}{M}^{{2}}+{A}{E}^{{2}}$ Pyth. theorem ${E}{M}$ $=\sqrt{{{A}{M}^{{2}}+{9}^{{2}}}}$  $={39.90537815637706}$ $\text{cm}$  In $\triangle{B}{E}{F}$, ${B}{E}^{{2}}$ $={E}{F}^{{2}}+{B}{F}^{{2}}$ Pyth. theorem ${B}{E}$ $=\sqrt{{{36}^{{2}}+{9}^{{2}}}}$  $={9}\sqrt{{{17}}}$ $\text{cm}$  In $\triangle{B}{E}{M}$, by the cosine formula, ${\cos}\angle{B}{M}{E}$ $=\dfrac{{{B}{M}^{{2}}+{E}{M}^{{2}}-{B}{E}^{{2}}}}{{{2}{\left({B}{M}\right)}{\left({E}{M}\right)}}}$  $={0.4472361746610752}$ $\angle{B}{M}{E}$ $={63.43350242369042}^{\circ}$

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