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Two ships are there in the sea on either side of a light house in such a way that the ships and the lighthouse are in the same straight line. The angles of depression of two ships as observed from the top of the lighthouse are 60° and 45°If the height of the lighthouse is 200m, what is the distance between the two ships?

Posted by ?‍? ??‍⚕? 6 years ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years ago

Let PM be the light house of height 200 m and let A and B be two ships on the either sides of lighthouse such that the angles of depression of A and B are $60^o$ and $45^o$ respectively.

Let, AM = x m and BM=y m

Then, $\angle XPB=\angle MBP=45^o$ [alternate angle]

and $\angle YPA=\angle MAP=60^o$ [alternate angles]

In right-angled $\triangle AMP,$ $\tan{60^o}=\frac{P}{B}=\frac{PM}{AM}$

$\Rightarrow \sqrt3=\frac{200}{x}$

$\Rightarrow x=\frac{200}{\sqrt3}\;m$

In right-angled $\triangle BMP,$ $\tan{45^o}=\frac{PM}{BM}$

$\Rightarrow 1=\frac{200}{y}$

$\Rightarrow y=200 \;m$

Now, distance between the two ships = AB = $x+y=200+\frac{200}{\sqrt3}=315.48\;m$

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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Posted by Pratibha Sahu 6 years ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years ago

Since, α and β are the zeroes of the quadratic polynomial
f(x) = x² - p (x+1) - c=x²-px-(p+c)
So A=1 B=-p,C=-(p+c)
Sum of the zeroes α + β = $-\frac BA$ =P
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(α + 1)(β + 1)
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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

2 answers

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

1 answers

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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CBSE > Class 10 > Mathematics

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Abhinav Yadav 6 years, 5 months ago

Start with ax^2 + bx + c=0 Divide the equation by a x^2 + bx/a + c/a = 0 Put c/a on other side x^2 + bx/a = -c/a Add (b/2a)2 to both sides x^2 + bx/a + (b/2a)^2 = -c/a + (b/2a)^2 The left hand side is now in the x2 + 2dx + d2 format, where "d" is "b/2a" So we can re-write it this way: "Complete the Square" (x+b/2a)^2 = -c/a + (b/2a)^2 Now x only appears once and we are making progress. Now Solve For "x" Now we just need to rearrange the equation to leave "x" on the left Start with (x+b/2a)^2 = -c/a + (b/2a)^2 Square root (x+b/2a) = (+-) sqrt(-c/a+(b/2a)^2) Move b/2a to right x = -b/2a (+-) sqrt(-c/a+(b/2a)^2) That is actually solved! But let's simplify it a bit: Multiply right by 2a/2a x = [ -b (+-) sqrt(-(2a)^2 c/a + (2a)^2(b/2a)^2) ] / 2a Simplify: x = [ -b (+-) sqrt(-4ac + b^2) ] / 2a

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CBSE > Class 10 > Mathematics