Question

The angles of elevation of an aeroplane, at an instant are found to be 60° and 30° from two places, on either side of the aeroplane. These two places are 10 km apart. Find the height at which the aeroplane is present at that instant.​

Answer 1

Answer:

Let PQ= h m be a plane

From right angled △PAQ

tan45

∘

=

AQ

PQ

∴1=

x

h

or x=h ......(i)

Again from right angled △PBQ

tan60

∘

=

QB

PQ

OR

3

=

2−X

H

From equation (i)

3

=

2−h

H

or 2

3

−

3

h=h

or h+

3

h=2

3

or h(1+

3

)=2

3

or h=

1+

3

2

3

=

1+

3

2

3

×

1−

3

1−

3

=

1−3

2

3

−6

=

−2

−2(3−

3

)

=(3−

3

)km

Answer 2

Refer to the attachment for diagram ✔️

Sólution:-

$➵tan {30}^{0} = \frac{h}{x + 10}$

$⟹x + 10 = \frac{h}{tan {30}^{0} }$

$⟹x + 10 = \sqrt{3} h \: \: \: \: \: (i)$

Again:-

$➵tan {60}^{0} = \frac{x}{h}$

$⟹x = \frac{h}{tan {60}^{0} } = \frac{h}{ \sqrt{3} } \: \: \: \: \: (ii)$

Subtracting both the equations:-

$⟹10 = \sqrt{3} h - \frac{h}{ \sqrt{3} }$

$⟹( \sqrt{3} - \frac{1}{ \sqrt{3} } )h = 10$

$⟹h = \frac{10}{ \sqrt{3} - \frac{1}{ \sqrt{3} } } = \frac{10 \sqrt{3} }{2} = 5 \sqrt{3}$

• Hence the height at which aeroplane ✈️ is present at that time is 5√3unit.

Hope you are helped ✔️