Question

An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. at the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. how far apart will be the two planes after 1 1/2 hours?​

Answer 1

Answer:300✓61km

Explanation:

brother distance equals to speed into time in a right angle triangle the square of hypotenuse is equal to the sum of the square of the other two sides AV the distance travelled by aeroplane travelling towards north

AB=1000 kilometre per hour×1×1/2

=1000×3/2 km

AB=1500km

bc is the distance travelled by another aeroplane travelling towards west.

=1200×3/2h

bc=1800km

now,In ∆ABC (phythagours theorem)

=(1500) whole square+(1800) whole square

=2250000+3240000

AC square=5490000

AC=✓5490000

=300✓61km

The distance between two planes after 1×2/2

he=300✓61km

Answer 2

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$\large \bf{ \orange{\underline{\underline{⚘Given:-}}}}$

$\bf{➢Speed \: Of \: north \: Flying \: aeroplane \: 1000 km/hr}$

$\bf{➢Speed \: of \: west \: Flying \: aeroplane= 1200 km/hr}$

$\large \bf{ \orange{\underline{\underline{⚘To \: Find:-}}}}$

$\bf{➢Distance \: between \: the \: two \: planes \: after \: 1.5 hours \: ,i.e., \: BC}$

$\large \bf{ \orange{\underline{\underline{⚘Solution:-}}}}$

$\bf {We \: know \: that }$

$\bf{ \pink➸ \: Speed = \dfrac{Distance}{Time} }$

$\bf{ \pink➸ \: Distance = Speed \times Time}$

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\tt{➭Speed = 1000 km/hr \tiny \green{(North \: Flying) }} \\ \tt{1200 Km/hr \tiny \green{(West \: Flying)}}\\ \\ \tt{➭Time= 1 \dfrac{1}{2} \tiny \green{(North \: Flying) }} \\ \tt{ 1 \dfrac{1}{2} \tiny \green{(West \: Flying)}}\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{➭Distance=AB=1000×1 \dfrac{1}{2}}\tiny \green{(North Flying) } \\ \tt{AC=1200×1 \dfrac{1}{2}}\tiny \green{(West \: Flying)}$

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$\bf{AB=1000×1 \dfrac{1}{2}}$

$\qquad \bf{ \red ➠\cancel {1000}} \times \dfrac{3}{ \cancel{2} }$

$\qquad \bf{ \red➠500 \times 3}$

$\qquad\frak{ \red➠1500 \: km}$

$\bf{AC=1200×1 \dfrac{1}{2}}$

$\qquad \bf{ \red ➠\cancel {1200}} \times \dfrac{3}{ \cancel{2} }$

$\qquad \bf{ \red➠600 \times 3}$

$\qquad \frak{ \red➠1800 \: km}$

$\bf{➢Now \: We \: have \: to \: find \: distance \: BC}$

$\bf{➢Since \: North \: and \: West \: are \: Perpendicular }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf{ \angle BAC=90 \degree}$

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$\bf{So \: \triangle ABC \: is \: a \: right \: angle \: triangle }$

$\bf{Using \: Pythagoras \: theorem \: in \: right \: angle \: triangle \: ABC}$

$\bf{(Hypotenuse)²=(Height)²+(Base)²}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ➭ \bf{(BC)²=(AB)²+(AC)²}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ➭ \bf{(BC)²=(1500)²+(1800)²}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ➭ \bf{(BC)²=2250000+3240000}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ➭ \bf{BC= \sqrt{5490000} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf{ \pink ➠ \sqrt{3 \times 3 \times 61 \times 100 \times 100} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf{ \pink ➠ \sqrt{ ( {3})^{2} \times (100)^{2} } \times \sqrt{61} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf{ \pink ➠ 3 \times 100 \times \sqrt{61} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar\underline{\boxed{\frak{ \orange{ 300 \sqrt{61} }}}}$

$\green\large\qquad \boxed{{\begin{array}{cc} \bf{Hence, \: Two \: planes \: would \: be \: 300 \sqrt{61} \: km} \\ \bf \: from \: each \: other.\end{array}}}$

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Concept Used Here

$\orange{\large\qquad \boxed{ \boxed{\begin{array}{cc} \\ \bf Pythagoras \: theorem\\ \\ \red{\boxed{ {\bf{(Hypotenuse)²=(Height)²+(Base)²}} } }\\ \\ \\ \\ \end{array}}}}$

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