Question

$\bigstar\:\:\underline{\red{\sf Question :}} \bigstar$
The angle of elevation of a jet fighter from a point O on the ground i 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet is flying at a speed of 720 km/hour, find the constant height at which the jet is flying. [Use √3=1.732.]
$\sf$
→ Spam answers will be deleted.
Don't be greedy for points ⚠️ ⚠️

→ Quality answers needed.​​

Answer 1

Answer:

$\huge{\sf\tt\large{\green {\underline {\underline{⚘\;solution:}}}}}$

⇰Let O be the point of observation on the ground OX.

⇰Let A and B be the two positions of the jet.

$\begin{gathered} \bf➠Then, \angle XOA=60⁰ \: and \: \angle \: XOB=30⁰ \\ \end{gathered}$

$\begin{gathered} \bf ➠Draw \: AL \perp \: OX \: and \: BM \perp \: OX \\ \end{gathered}$

$\begin{gathered} \bf \:⇰ Let \: AL=BM=h \: meters \: \end{gathered}$

$\begin{gathered} \bf \: ⇰ \: Speed \: of \: Jet = 720 \: km/hr\\ \end{gathered}$

$\begin{gathered} \bf = \huge( \small \: 720 \times \frac{5}{18} \huge) \small \: m / s\\ \end{gathered}$

$\begin{gathered} \bf \:➥ Time \: taken \: to \: cover \: the \: distance \: AB=15 \: sec \end{gathered}$

$\begin{gathered} \bf➥ Distance \: covered =(speed×time) \end{gathered}$

$\begin{gathered} \: \: \: \red{ \bf = (200 \times 15)m = 3000m} \\ \\ \end{gathered}$

$\begin{gathered} \bf \therefore LM=AB=3000m \end{gathered}$

$\begin{gathered}{ \sf\bf \: ⇰ \: Let \: OL} = \sf \: x \: \bf m \end{gathered}$

$\begin{gathered} \bf➥From \: right \: \triangle OLA, \: we \: have: \\ \end{gathered}$

$\begin{gathered} \bf \frac{OL}{AL} =cot \: 60⁰= \frac{1}{ \sqrt{3} } \implies \frac{x}{h} = \frac{1}{ \sqrt{3} } \\ \\ \end{gathered}$

$\begin{gathered} \bf \implies \: x = \frac{h}{ \sqrt{3} } \: \: \: \: \: \: \: ....(1) \\ \\ \end{gathered}$

$\begin{gathered} \bf \:➥ From \: right \: \triangle \: OMB,we \: have: \\ \end{gathered}$

$\begin{gathered} \bf\frac{OM}{BM} =cot30⁰= \sqrt{3} \\ \\ \end{gathered}$

$\begin{gathered} \bf \implies \: \frac{x + 3000}{h} = \sqrt{3} \\ \\ \end{gathered}$

$\bf \sf\boxed{ \bigg[ \sf➯OM=OL+LM=OL+AB=(x+3000)m \: and \: BM=h \: m \bigg]}$

$\begin{gathered} \bf \implies \: x + 3000 = \sqrt{3} h \\ \end{gathered}$

$\begin{gathered} \bf \implies \: x = ( \sqrt{3} h - 3000) \: \: \: \: \: \: ....(2) \\ \end{gathered}$

⇰ Equating the value of x from (1) and (2),we get;

$\begin{gathered} \bf \: \sqrt{3} h - 3000 = \frac{h}{ \sqrt{3} } \\ \\ \end{gathered}$

$\begin{gathered} \implies \bf \: 3h - 3000 \sqrt{3} = h \\ \end{gathered}$

$\begin{gathered} \bf \implies \: 2h = (3000 \times \sqrt{3} ) = (3000 \times 1.762) \\ \end{gathered}$

$\begin{gathered} \bf \implies \: h = (3000 \times 0.866) = 2598 \\ \\ \end{gathered}$

$\sf Hence \: ,the \: required\: height \: is \: \boxed{ \boxed{ \bf\red{2598 \: m}}}$

☆To View full answer drag your screen ⇰

Answer 2

Hence, the required height is 2598 m.

$\sf$

Answer refers in attachment

It helps uhh. . . ♡