Saturday

February 13, 2016
Posted by **Crystal** on Saturday, November 24, 2012 at 8:18pm.

The answer is 150 km, please explain to me how

4) from a boat on the water, the angle of elevation at the top of the cliff is 31 degrees. From a point 300 m closer, the angle of elevation is 33 degrees. Find the height of the cliff.

Answer is 2411 m

5) from the top of a cliff 185 m high, the angles of depression of two channel buoys in the same line of sight on the water are 13 degrees and 15 degrees. How far apart are the buoys?

The answer is 111 m.

I'm having a lot of trouble figuring out these questions. When I try to do them, I feel like there's missing information. Or I'm misinterpreting the questions. Please help

- Math -
**Damon**, Saturday, November 24, 2012 at 8:53pmSKETCH the problem

draw altitude h down from C to line AB at D

call AD x

then DB is 350 -x

Now look at right triangle ACD

tan 32 = h/x = .625

so

x = h/.625 = 1.6 h

Now triangle BCD

tan 54 = h/(350-x)

1.38 = h/(350 -1.6 h)

482 - 2.20 h = h

3.2 h = 482

h = 151 meters

- Math -
**Damon**, Saturday, November 24, 2012 at 8:54pmIf you follow that you can do the other two.

- Math -
**Damon**, Saturday, November 24, 2012 at 9:00pmSorry, kilometers, not meters

Look down at the bottom of the page for related problems solved. Look particularly at this one:

http://www.jiskha.com/display.cgi?id=1279657357

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**Crystal**, Saturday, November 24, 2012 at 10:40pmHow did you get x = h/.625 = 1.6 h ? How did you arrive at 1.6?

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**Damon**, Saturday, November 24, 2012 at 10:58pm1/.625 = 1.6

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**Damon**, Saturday, November 24, 2012 at 11:13pmtan 32 = h/x

so

x tan 32 = h

so

x = h/tan 32

so

x = h/.625

but 1/.625 = 1.6

so

x = 1.6 h

- Math -
**Crystal**, Saturday, November 24, 2012 at 11:43pmThanks! I don't know how I missed that.

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**Crystal**, Sunday, November 25, 2012 at 12:10amokay so I think I'm still kind of hopeless... my solution to the second question is way off, but here's what I did:

right triangle ABC. (facing like this: >) so that AB is the hypotenuse, and I labelled AC as h for the height of the cliff. I drew a dotted line from point A to about the middle of line BC and called that point D.

Made angle ABD 31 degrees, angle ADC 33 degrees, line BD x+300, line DC x.

for triangle ADC:

tan33=h/x

x=h/tan33

x=h/0.65

for triangle ABC:

tan 31=h/x+300+x

0.6 = h all over h/0.65 +300 + h/0.65

0.6 =h x (1.3/2h) + 300

0.6 - 300 = 1.3/h

-299.4 = 1.3/h

h = 1.3/299.4

h = 0.004 .....

Yeah, I guess I really don't get this... *sigh*

- Math -
**Damon**, Sunday, November 25, 2012 at 6:38amIt was midnight. I went to bed.

Made angle ABD 31 degrees, angle ADC 33 degrees, line BD x+300, line DC x.

for triangle ADC:

tan33=h/x

x=h/tan33

x=h/0.65

for triangle ABC:

tan 31 = h/(x+300)

.6 = h/(x+300)

.6 x + 180 = h

.6(h/.65) + 180 = h

.92 h + 180 = h

180 = .08 h

h = 2250

need to carry more significant figures to get a more accurate answer

- Math -
**Crystal**, Sunday, November 25, 2012 at 9:05amCompletely understandable, thank you so much for checking back though!