### angle of elevation

Question Sample Titled 'angle of elevation'

In the figure, ${O}{T}$ is a vertical tower on a horizontal ground. ${P}$ and ${Q}$ are two points on the ground such that the angles of elevation of ${T}$ from ${P}$ and ${Q}$ are ${46}^{\circ}$ and ${34}^{\circ}$ respectively. It is given that $\angle{P}{O}{Q}={100}^{\circ}$ and ${P}{Q}={890}$ $\text{m}$.

${P}$${Q}$${T}$${O}$${100}^{\circ}$${890}$ $\text{m}$${46}^{\circ}$${34}^{\circ}$

 (a) (i) Let ${h}$ $\text{m}$ be the height of the tower. Express ${O}{P}$ and ${O}{Q}$ in terms of ${h}$. (ii) Hence, find ${h}$.  (4 marks) (b) Suppose Maria walks along ${P}{Q}$. She claims that the maximum angle of elevation of ${T}$ from her occurs at the mid-point of ${P}{Q}$. Do you agree? Explain your answer. (3 marks)

 (ai) ${O}{P}=\dfrac{{h}}{{{\tan{{46}}}^{\circ}}}$ $\text{m}$ 1A ${O}{Q}=\dfrac{{h}}{{{\tan{{34}}}^{\circ}}}$ $\text{m}$ 1A (aii) Consider $\triangle{O}{P}{Q}.$ By the cosine formula, ${O}{P}^{{2}}+{O}{Q}^{{2}}-{2}{\left({O}{P}\right)}{\left({O}{Q}\right)}{\cos}\angle{P}{O}{Q}$ $={P}{Q}^{{2}}$ ${\left(\dfrac{{h}}{{{\tan{{46}}}^{\circ}}}\right)}^{{2}}+{\left(\dfrac{{h}}{{{\tan{{34}}}^{\circ}}}\right)}^{{2}}-{2}{\left(\dfrac{{h}}{{{\tan{{46}}}^{\circ}}}\right)}{\left(\dfrac{{h}}{{{\tan{{34}}}^{\circ}}}\right)}{{\cos{{100}}}^{\circ}}$ $={890}^{{2}}$ 1M Solving, we have  ${h}={467}$ (cor. to 3 sig. fig.) 1A

 (b) Let ${M}$ be a point on ${P}{Q}$ such that the angle of elevation of ${T}$ from ${M}$ is maximum. The angle of elevation is maximum only when ${O}{M}$ is the shortest, i.e. ${O}{M}\bot{P}{Q}$. 1M If ${M}$ is the mid-point of ${P}{Q}$, as ${O}{M}\bot{P}{Q}$, then, $\triangle{O}{M}{P}\stackrel{\sim}{=}\triangle{O}{M}{Q}$. SAS 1M Then, ${O}{P}={O}{Q}$. corr. sides, ≅△s However, as ${{\tan{{46}}}^{\circ}\ne}{{\tan{{34}}}^{\circ}}$, ${O}{P}\ne{O}{Q}$ $\therefore{M}$ must not be the mid-point of ${P}{Q}$. $\therefore$She is not correct. 1A

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