Question

A man on a cliff observes a boat at an angle of depression of 30deg which is approaching the shore to the pointimmediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boatis found to be 60deg. Find the time taken by the boat to reach the shore. plz answer this......anyone

Answer 1

Hi ,

Draw a rough diagram to the data

given in the question,

|A
|
|
|
| h
|
| 60
|________|_____30
B <----y----->C.<-x--> D

Join A to C and D

Height of the cliff = AB = h

BC = y

Distance travelled in 6 mi

= CD = x

i ) in traingle ADB ,

Angle ABD = 90

Tan 30 = h / ( x + y )

1/ √3 = h / ( x + y )

( x + y )/( √3 ) = h -----( 1 )

ii ) In triangle ABC

Angle B = 90

Tan 60 = h / y

√3 = h / y

( √ 3 ) y = h ------( 2 )

From ( 1 ) and ( 2 )

We observe that

( 1 ) = ( 2 )

( x + y ) / (√ 3 ) = ( √3 ) y

x + y = 3y

x = 3y - y

x = 2y -----( 3 )

iii ) according to the problem given

distance ' x ' is travelled in

6 minutes

How much time taken to

travel distance ( x + y ) = ?

Time taken

= [ 6 × ( x + y ) ] / x

= [ 6 × ( 2y + y ) ] / 2y

{ since x = 2y from ( 3 ) ]

= [ 6 × 3y ] / 2y

= 18 y / 2y

= 9 minutes

Therefore ,

In 9 minuates the boat

reaches the shores .

I hope this helps you.

:)

Answer 2

ANSWER:

_____________________________

Let OA be the cliff.

P be the initial position of the boat when the angle of depression is 30°

After 6 mins the boat reaches to Q such that the angle of depression at Q is 60°.

Let PQ=x metres.

Now we have two triangles.

In Δ's PQA and QOA we have,

tan 30°= OA/OP and

tan 60°=OA/OQ

$= > \frac{1}{ \sqrt{3} } = \frac{AO}{OP} \: \: \: \: and \\ \\ = > \sqrt{3} = \frac{OA}{OQ} \\ \\ = > oa = \frac{OP}{ \sqrt{3} } \: and \: \: OA = \sqrt{3} \: OQ \\ \\ = > \frac{op}{ \sqrt{3} } = \sqrt{3} \: \: OQ \: \\ \\ = > OP = 3 \: OQ \\ \\ = > PQ = OP - OQ = OP - \frac{OP}{3} = \frac{2}{3} \: OP .......(OQ = \frac{1}{3} OP)$

Let the speed of boat be v meter/minute

Then,

PQ=Distance travelled by the boat in 6 minutes.

=>PQ =6v

$\\ = > \frac{2}{3} (OP) = 6v............( PQ = \frac{2}{3}OP) \\ \\ = > OP = 9v \\ \\ now \\ \\ time \: taken \: by \: the \: boat \: to \: reach \: the \: shore \: is \: given \: by \\ \\ t = \frac{distance}{time} \\ \\ t = \frac{op}{v} \\ \\ = > t = \frac{9v}{v} \\ \\ t = 9 \: minutes \\ \\ = > time \: taken \: by \: the \: boat \: to \: reach \: the \: shore \: \: = 9 \: minutes$