Download Section 5.7 (part 1): Solving Right Triangles, SAS, More

Section 5.7 (part 1): Solving Right Triangles, SAS, More SOHCAHTOA
In this section, we have two main goals. We first show how we can find all sides and angles of right triangles
in some cases. Next, we’ll look at harder applications of breaking shapes into right triangles. Along the way,
we’ll get a useful SAS Area Formula for triangles where you know two sides and an angle in between.
Techniques for Solving Triangles, part 1
Standard notation for ∆ABC:
• Angles: α, β, and γ (at A, B, C respectively)
– Usually, for right triangles, γ = 90◦ = π/2.
• Sides: a, b, c (opposite A, B, C)
Solving the triangle means finding all the sides
and angles. With a right triangle, we pick one of the
following approaches based on what we already know:
•
•
•
•
If
If
If
If
you
you
you
you
know
know
know
know
two
two
one
two
sides and want the third: Pythagoras.
angles and want the third: the angles add to 180◦ .
side and one acute angle, and you want another side: SOHCAHTOA.
sides and want an acute angle:
SOHCAHTOA gives you a trig equation with the angle. Get the acute solution! (Special angle?)
RECOMMENDATION: Label your steps on the picture; this helps you keep track of what you’ve already
found and what you need to find next. It also means your scratchwork can just indicate answers to old steps
without recopying the entire value on paper.
Ex 1: Solve ∆ABC given:
√
√
(b) γ = 90◦ , b = 4 2, c = 8 2
(a) γ = 90◦ , α = 30◦ , b = 25
SAS Area Formula
SAS stands for “Side, Angle, Side”... this occurs
when you know two sides and the included angle between them. A typical SAS scenario is shown on the
right, where b, c, and α are known.
The base here is c. You find the height by using b,
α, and SOHCAHTOA to get h = b sin(α). This gives
you the following:
SAS Area Formula: Area = 12 cb sin(α) i.e. (1/2)(product of two sides)sin(angle between)
Ex 2: Find the area of a parallelogram which has sides of length a = 13.7 and b = 15.6 feet and one angle
of measure θ = 38◦ .
HINT: Draw a diagonal to break this into SAS triangles.
One place we’ll frequently see SAS get used is in pictures of a regular polygon along with a circle. If you
have a regular n-gon, then you can draw n total radii from the center of the polygon to get n congruent “central
triangles”. (Think of the radii as being like spokes on a wheel.) The n total central angles add to 360◦ , so each
triangle has an angle of (360/n) degrees. When you know the radius, you get n SAS triangles!
Ex 3: A regular octagon is inscribed in a circle of radius 19. Determine the area of the region in the circle
but outside the octagon.
GETTING STARTED: Break the octagon into 8 central triangles. Each has two “spokes” of length 19 and
a central angle of (360/8) = 45 degrees.
Ex 4: An equilateral triangle is inscribed in a circle of radius 7. Determine the area of the region in the
circle but outside the triangle.
NOTE: Your equilateral triangle’s sides are not length 7. (Why not?) Instead, split this triangle into three
central triangles with radii 7 and 120◦ central angles.
Miscellaneous Problems
Here are some problems using the ideas we’ve introduced. For many of these problems, you want to break
up shapes into right triangles (by drawing heights, for instance). In some cases, it’s easier to make triangles
which use the SAS area formula.
Ex 5: An isosceles triangle has an angle of t = 37◦ between two sides of length 18. Find the area of the
triangle and the length of the third side.
HINT: Draw a height for this triangle, so that you obtain two right triangles.
Ex 6: A circle is inscribed in a regular pentagon
with side length 11 inches, as shown in the figure. OA
is a radius of the circle.
(a) Find angle AOB in degrees.
HINT: Imagine making copies of ∆AOB all
around the pentagon. How many triangles cover
each side of the pentagon? How many total triangles would you have?
(b) Find the length of OA in inches.
NOTE: OA is NOT 11... and furthermore, OA 6=
OB. (Why is OB NOT a radius?) Instead, start
with the fact that AB is 11/2.
(c) Find the area of the circle in square inches.
Ex 7 (waterslide): The figure shows a design of
a water slide. Find the total length of the slide, if
α = 25◦ , β = 31◦ , a = 35 ft, and b = 190 ft.
ORGANIZING THIS:
1. You can break the region under the slide into two
rectangles and two right triangles.
2. Use SOHCAHTOA to get the lengths of the
slanted parts of the slide.
3. To get the length of the flat part, it’s easier to figure out the bases of your triangles and subtract
them from b.
Elevation and Depression (part 1)
One handy use of angles is for elevation or depression angles. When a side is angled upwards, the elevation angle is the angle made with the horizontal at
the ground, whereas the depression angle is the angle made with the horizontal at the top. In the figure,
α is elevation, and β is depression.
Note that α and β are the same measure! They are
alternate interior angles.
Ex 8: From a point 25 meters above level ground, a surveyor measures the angle of depression of an object
on the ground at 64◦ . Approximate the distance from the object to the point on the ground directly beneath
the surveyor.
Ex 9: A rocket climbs at a 70◦ angle through 30, 000 feet. Determine the altitude of the rocket.
NOTE: The altitude is the height of the rocket. The 30, 000 feet measurement represents the hypotenuse!
(After all, the rocket is traveling diagonally!)
We’ll do a couple more examples with this next class.