Once you've learned about trigonometric
ratios (and their inverses), you can solve triangles. Naturally, many
of these triangles will be presented in the context of word problems.
A good first step, after reading the entire exercise, is to draw a right
triangle and try to figure out how to label it. Once you've got a helpful
diagram, the math is usually pretty straightforward.
A six-meter-long ladder
leans against a building. If the ladder makes an angle of 60° with
the ground, how far up the wall does the ladder reach? How far from
the wall is the base of the ladder? Round your answers to two decimal
places, as needed.
First, I'll draw a picture. It
doesn't have to be "good"; it just needs to be clear
enough that I can keep track of what I'm doing.
My picture is at right:
I need to find the height h.
Since they've given me an angle measure and "opposite"
and the hypotenuse for this angle, I'll use the sine ratio for
finding the height:
= h = 3sqrt
Plugging this into my calculator,
I get an approximate value of 5.196152423.
For the base, I'll use the cosine ratio:
cos(60°) = b/6
= b = 3
The ladder reaches
about 5.20 meters up the wall,
and its base is 3 meters from the
Note: Unless you are told to give your
answer in decimal form, or to round, or in some other way not to
give an "exact" answer, you should probably assume that the
"exact" form is what they're wanting. For instance, if they
hadn't told me to round in the exercise above, my value for the height
should have been the value with the radical.
A five-meter-long ladder
leans against a wall, with the top of the ladder being four meters above
the ground. What is the approximate angle that the ladder makes with
As usual, I'll start with a picture:
They've given me the "opposite"
and the hypotenuse, and asked me for the angle value. For this, I'll
need to use inverse trig ratios.
sin(α) = 4/5
= sin–1(4/5) = 53.13010235...
The angle the ladder makes with the ground
is about 53°.
You use a transit to
measure the angle of the sun in the sky; the sun fills 34' of arc. Assuming
the sun is 92,919,800 miles away, find the diameter of the sun. Round
your answer to the nearest mile.
First, I'll draw a picture:
Hmm... This "ice-cream cone"
picture doesn't give me much to work with, and there's no right
The two lines along the side of
my triangle measure the lines of sight from Earth to the sides
of the Sun. What if I add another line, being the direct line
from Earth to the center of the Sun?
Now that I've got this added line,
I have a right triangle; two right triangles, actually, but I
only need one.
Because the Sun is so far away,
obviously this picture is not "to scale", and I'll have
to ignore the fact that the distance measured doesn't quite match
the picture. But this should be good enough.
measure, "thirty-four arc minutes", is equal to
34/60 degrees. Dividing this in half is how I got 17/60 of a degree
for the smaller angle.)
tan(17/60°) = b/92919800
= b = 459501.4065...
This is just half the width; carrying
the calculations in my calculator (to minimize round-off error), I get
a value of 919002.8129.
This is higher than the actual diameter, which is closer to 864,900 miles, but this value will suffice for the purposes of this exercise.
The diameter is about
A private plane flies
1.3 hours at 110 mph on a bearing of 40°. Then it turns and continues
another 1.5 hours at the same speed, but on a bearing of 130°. At
the end of this time, how far is the plane from its starting point?
What is its bearing from that starting point?
The bearings tell me the angles from "due north", in a clockwise direction.
Since 130 – 40 = 90, these two bearings will give me a right triangle.
From the times and rates, I can find the distances:
1.3 × 110 = 143
× 110 = 165
Now that I have the lengths of the two
legs, I can set up a triangle: