04Felevation-depression

toc = Angles of Elevation & Depression =

Literacy
To answer word problems in trigonometry, there are two extra words for you to learn:

Angle of Elevation

An ** angle of elevation ** is the angle __above__ a horizontal line.
 * {Elevate means "to lift up". An elevator (or lift) takes you __up__ inside a building}
 * The horizontal line starts at the observer

Angle of Depression

An ** angle of depression ** is the angle __below__ a horizontal line.
 * {Depress means "to press down". A person who is depressed is feeling down.}
 * The horizontal line starts at the observer

"** Line of Sight **" is a straight line from the observer to the object they are looking at.
 * It will usually be the __hypotenuse__ of the right angled triangle in the diagram.

Word Problems
When answering word problems
 * draw a diagram
 * include a horizontal line from the observer
 * then a vertical line to the object being observed
 * this should create a right angled triangle with the line of sight as the hypotenuse
 * the angle of elevation or depression is the angle between the __horizontal line__ and the line of sight
 * use a pronumeral for the length or angle that is to be found
 * use trigonometry to find the value of the pronumeral
 * express your answer in terms of the original problem. Don't forget to include units.

** Example 1 **
Jo is flying a kite so that the string forms an angle of elevation of 56º. If the string is 15m long, how high is the kite?

__**Solution:**__

{Draw Diagram} Use h = height of kite

{From the diagram, HYP = 15m, OPP = h, so use __**sin**__}

math \\ . \qquad \sin \big( \theta \big) = \dfrac{\text{OPP}}{\text{HYP}} \\ \\ \\ . \qquad \sin \big( 56^\circ \big) = \dfrac{h}{15} \qquad \{ \times 15 \} \\ \\ . \qquad h = 15 \times \sin \big( 56^\circ \big) \\ \\ . \qquad h = 12.4 \text{ m} math

The kite is 12.4m above the ground.

** Example 2 **
A fire spotter sits in a tower 200m high and observes smoke at an angle of depression of 11º. Assume the ground is horizontal, how far is the fire from the base of the tower?

__**Solution:**__

{Draw a diagram} Use d = distance to fire

{From the diagram, OPP = 200, ADJ = d, so use __**tan**__} math \\ . \qquad \tan \big( \theta \big) = \dfrac{\text{OPP}}{\text{ADJ}} \\ \\ \\ . \qquad \tan \big( 11^\circ \big) = \dfrac{200}{d} \qquad \{ \times d \} \; and \; \{ \div \tan(11^\circ) \} \\ \\ . \qquad d = \dfrac{200}{ \tan \big( 11^\circ \big) } \\ \\ . \qquad h = 1028.9 \text{ m} math

The fire is 1028.9 m (1.03 km) from the base of the tower.

** Example 3 **
A wire 31m long is attached to the top of a 28m vertical flag pole. To help hold the pole upright, the wire is stretched taught and attached to the ground. What will be the angle of elevation of the wire at the point where it is attached to the ground?

__**Solution:**__

{Draw a diagram} Use a = angle of elevation

{From the diagram, OPP = 28, HYP = 31, so use __**sin**__}

math \\ . \qquad \sin \big( \theta \big) = \dfrac{\text{OPP}}{\text{HYP}} \\ \\ \\ . \qquad \sin \big( \alpha \big) = \dfrac{28}{31} \\ \\ . \qquad \alpha = \sin^{-1} \left( \dfrac{28}{31} \right) \\ \\ . \qquad \alpha = 64.6^\circ math

The angle of elevation is 64.6º above the horizontal.

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