Angle of depression.

**Mr Rayon** · June 23rd 2010, 02:41 PM

Kath stands at the top of a vertical cliff. She sights a ship on the ocean at an angle of depression of 7 degrees. The ship moves directly away from where Kath is standing, until after it has travelled 1100 m, the angle of depression is now 5 degrees.
a) Use the sine rule to find the direct (line of sight) distance from Kath to the ship at its second location, correct to the nearest metre.
b) Hence or otherwise find, correct to nearest metre:
i) the height of Kath above the ocean
ii) the distance of the ship from the base of the cliff before it travelled the 1100 m.

Please show any working out. I would love to see some help regarding these questions.

**bigwave** · June 23rd 2010, 02:58 PM

first by geometry get the interior angles of the triangle then use the law of sines to get the sides.
the interior angles should be (that is if I see this correctly) are $2^o$ $5^o$ and $173^o$

since the "direct(line of sight) distance" is opposite of $173^o$ and $1100m$ is opposite of $2^o$ we can use the law of sines to get the distance
now solve for $x$ which is the distance of the line of sight
$\frac{\sin{173^o}}{x} = \frac{\sin{2^o}}{1100}$

can you get the rest of it.....

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