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Mathematics 10 Online
OpenStudy (anonymous):

A man Standing on top of A cliff 50m high is in line with two buoys whose angles of depression are 18degree and 20degree. Calculate the distance between the buoys.

OpenStudy (dumbcow):

Use tan function: tan = opp/adj Let x be distance from cliff to 1st buoy \[\tan 20 = \frac{x}{50}\] Let y be distance from cliff to 2nd buoy \[\tan 18 = \frac{y}{50}\] Distance between buoys is y-x

OpenStudy (anonymous):

Well, the first buoy is \[50\tan 18\]50tan(18) meters away (basic right angled triangle) The second one is 50tan(20) meters away. We calculate these and get a difference.

OpenStudy (anonymous):

The answer is 16.5m

OpenStudy (dumbcow):

thats the idea..my angles may be off

OpenStudy (dumbcow):

depression meaning angle of line from top of cliff hmm that means the compliment is the angle you want to take tangent of

OpenStudy (anonymous):

Ah right.

OpenStudy (dumbcow):

\[1st = 50*\tan 70 =137.37\] \[2nd = 50*\tan 72 = 153.88\] \[153.88-137.37 = 16.51\]

OpenStudy (anonymous):

tan (90-20) tan(90-18)

OpenStudy (anonymous):

those should be the angle you should use ..

OpenStudy (anonymous):

Thank

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