Download NOTES Afm (ASA Triangle)(Ssa Triangle) = C

NOTES
Afm
-Oblique Triangle- A triangle that does not contain a right angle.
(oblique triangle)
-Solving a oblique triangle means finding the lengths of its sides and the
measurement of its angles.
-The three known measurements can be abbreviated using SAA (a side
and two angles are known) or ASA (two angles and the side between them
are known)
-Law of sines
-If A,B and C are the measures of the angles of a triangle, and a,b and c are the lengths of the
sides opposite these angles then,
a/SinA=b/SinB=c/SinC
-The ratio of the length of the side of any triangle to the sine of the angle opposite that side is
the same for all three sides of the triangle.
c
a
AD
(ASA Triangle)(Ssa Triangle)
-In an SSA situation, it is not necessary to draw an accurate sketch like those shown in
the box.
-The law of sines determines the number of triangles, if any, and gives the solution for
each triangle.
𝐴=𝐶
𝐴 = 56
𝐴 = 56𝑆𝑖𝑛46 𝐴 ≈ 45 𝐼𝑛𝑐ℎ𝑒𝑠
𝑆𝑖𝑛𝐴 = 𝑆𝑖𝑛𝐶 𝑆𝑖𝑛 46 = 𝑆𝑖𝑛 63
𝑆𝑖𝑛63
-Solving an ASA Triangle using the Law of sines
A+B+C=180(the sum of the measures of a triangle interior angle is 180)
50+B+33.5 =180(A=50 / C=33.5 )
83.5+B=180 ( Add )
B=96.5 ( subtract 83.5 from both sides )
c
33.5
a
B=76
50
B
(The area of an oblique triangle)
A formula for the area of an oblique triangle triangle can be obtained using
the procedure a formula the law of sines. We draw an altitude for length h
from one of the vertices of the triangle.
Sine of angle A,
!""#$%&'
!"#$%&'()&
, in the right triangle ACD
!
Sin A= ! or h=b sin A.
Area=½ cb sin A or ½ bc sin A, indicates that the area of the triangle is one
half the product of b and c times the sine of their included angle. If we draw
altitudes from the other two vertices, you can use any two sides of the
computed area.
The area of a triangle is ½ the product of any side and the altitude drawn to
that side.
Area= ½ ch = ½ cb sin A.
C
b
A
D
If you add either angle to the given
angle , 40 , the sum does not exceed 180 Thus, these are two triangles
with the given conditions. Triangle AB1 C1 and Ab2C2.
To find b equation using the law of sines, you need to know the side
and an angle opposite of that side.
If a=90-50=40
B=90-36=54
C=180-A-B=180-40-54=86
(You may have seen how the trigonometry of right triangles and how
they can be used to solve many different kinds of applied problems)
The law of sine enables you to work with triangles that are not right
triangles.As a result , this law can be used to solve problems
involving surveying, engineering, astronomy, navigation and the
environment.
b
54
4
C=20
THE LAW OF SINES
If A,B and C are the measures of the angles of a triangle and a,b, and
c are the lengths of the sides opposite these angles , then
a^2 =b^2 + C^2 - 2bc cos A
b^2 =a^2 +c^2- 2ac cos B
C^2 = a^2 +b^2 - 2ab cos C
The square of a side of a triangle equals the sum of the squares of the
other two sides minus twice their product times the cosine of their
included angle.
Example- To prove the law of cosines we place triangle ABC in a
rectangular coordinate system. The vertex A is at the origin and slide
c lies along the positive x axis the coordinates of C are (x,y) Using the
right triangle that contains angle A , we apply the definitions of the
cosine and the sine.
( EXAMPLE) find the area of a triangle having two sides of lengths 24
meters and 10 meters and an included angle of 62.
(SOLUTION) Its area is half the product of the lengths of the two sides
times the sine of the included angle.
Area=½ (24)(10)(sin 62) = 106
a^2 =b^2+c^2 - 2b cos A (apply the law of cosines to a)
a^2=20^2 + 30^2-2(20)(30) cos 60 (b =20 , c =30 and A=60
=400 + 900- 1200 (0.5) (perform the indicated operations)
=700
a=700
a=700=26 ( Take the square root of both sides and solve for a)
*use the law of sines to find the angle opposite the shorter of two
given sides. The angle is always acute.