of Elevation / Inclination and
of Depression / Declination
Angles of elevation or inclination are
angles above the horizontal, like looking up from ground level toward
the top of a flagpole. Angles of depression or declination are angles
below the horizontal, like looking down from your window to the base of
the building in the next lot. Whenever you have one of these angles, you
should immediately start picturing how a right triangle will fit into
Driving along a straight
flat stretch of Arizona highway, you spot a particularly tall saguaro
("suh-WARH-oh") cactus right next to a mile marker. Watching
your odometer, you pull over exactly two-tenths of a mile down the road.
Retrieving your son's theodolite from the trunk, you measure the angle
of elevation from your position to the top of the saguaro as 2.4°.
Accurate to the nearest whole number, how tall is the cactus?
Two-tenths of a mile is 0.2×5280 feet = 1056 feet, so this is my horizontal distance. I need to find the height h of the cactus. So I draw a right triangle and label everything I know:
The scale is not important; I'm
not bothering to get the angle "right". I'm using the
drawing as a way to keep track of information; the particular
size is irrelevant.
What is relevant is that
I have "opposite" and "adjacent" and an angle
measure. This means I can create and solve an equation:
h/1056 = tan(2.4°) h =
1056×tan(2.4°) = 44.25951345...
To the nearest foot, the
saguaro is 44 feet tall.
You were flying a kite
on a bluff, but you managed somehow to dump your kite into the lake
below. You know that you've given out 325 feet of string. A surveyor tells you that the angle of declination from
your position to the kite is 15°.
How high is the bluff where you and the surveyor are standing?
First, I draw my triangle:
The horizontal line across the
top is the line from which the angle of depression is measured.
But by nature of parallel lines, the same angle is in the bottom
triangle. I can "see" the trig ratios more easily in
the bottom triangle, and the height is a bit more obvious. So
I'll use this part of the drawing.
I have "opposite",
hypotenuse, and an angle, so I'll use the sine ratio to find the
stands on a hill 100 m above sea level. If ∠ACD measures 60° and ∠BCD is 30°,
find the height of the lighthouse.
I'm going to have to work this exercise
in steps. I can't find the height of the tower, AB,
until I have the length of the base CD.
(Think of D as being moved to the right, to meet the continuation of AB,
forming a right triangle.) For this computation, I'll use the height
of the hill.
To minimize round-off error, I'll use
all the digits from my calculator in my computations, and try to "carry"
the computations in my calculator the whole way..
Now that I have the length of the base,
I can find the total height, using the angle that measures the the elevation
from sea level to the top of the tower.
h/173.2050808 = tan(60°) h = 173.2050808×tan(60°)
Excellent! By keeping all the digits
and carrying the computations in my calculator, I got an exact answer.
No rounding! But I do need to subtract, because "300"
is the height from the water to the top of the tower. The first hundred
meters of this total height is hill, so: