Exam 11: Introduction to Trigonometry
Exam 1: Line and Angle Relationships13 Questions
Exam 2: Parallel Lines13 Questions
Exam 3: Triangles16 Questions
Exam 4: Quadrilaterals14 Questions
Exam 5: Similar Triangles12 Questions
Exam 6: Circles10 Questions
Exam 7: Locus and Concurrence4 Questions
Exam 8: Areas of Polygons and Circles5 Questions
Exam 9: Surfaces and Solids4 Questions
Exam 10: Analytical Geometry8 Questions
Exam 11: Introduction to Trigonometry4 Questions
Exam 12: Geometry Problems: Complementary Angles, Collinear Points, and Similar Triangles916 Questions
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-Use the drawings provided to prove the following theorem.
"The area of an acute triangle equals one-half the product of the lengths of two sides of a triangle and the sine of the included angle."
Given: Acute
Prove: 



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Correct Answer:
The area of is given by
. Considering the auxiliary altitude
from vertex B
to side , we see that
is a right angle. Then
in right triangle
.
From , it follows that
. By substitution, the area formula
becomes
so we have
.
-For the right triangle
with right
, prove that
.




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Correct Answer:
By definition, ,
, and
. Then
, which can also be written
. Then
. Thus,
.
-For the right triangle
with right
, prove that
.




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Correct Answer:
In right triangle ,
. Because
is the reciprocal of
, we know that
. Now
, so
or
.
Then . Because
is a right triangle with right angle at vertex C,
. Then
; that is,
.
-For the right triangle
with right
, prove that
. Note that
and
are the same and that
as shown in
; also,
and
have the
same meaning.










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