Download Solving Triangles. For a triangle with angles A, B, C with sides of

Solving Triangles.
For a triangle with angles A, B, C with sides of length a, b, c where a is across from(opposite) A, b is
across from(opposite) B, and c is across from(opposite) C
The law of sines states that:
sin ‫ܣ‬
sin ‫ܤ‬
sin ‫ܥ‬
=
=
ܽ
ܾ
ܿ
It can also be used in this form:
ܽ
ܾ
ܿ
=
=
sin ‫ܣ‬
sin ‫ܣ‬
sin ‫ܥ‬
The law of cosines states that:
ܿ ଶ = ܽଶ + ܾ ଶ − 2ܾܽ cos ‫ܥ‬
ܾ ଶ = ܽଶ + ܿ ଶ − 2ܽܿ cos ‫ܤ‬
ܽଶ = ܾ ଶ + ܿ ଶ − 2ܾܿ cos ‫ܣ‬
It can also be used in this form:
cos ‫= ܥ‬
ܽଶ + ܾ ଶ − ܿ ଶ
2ܾܽ
cos ‫= ܤ‬
ܽଶ + ܿ ଶ − ܾ ଶ
2ܽܿ
cos ‫= ܣ‬
ܾ ଶ + ܿ ଶ − ܽଶ
2ܾܿ
What do we mean by “Solving a triangle?”.
A triangle has six properties to determine: The measures of three angles, and the lengths of three sides.
When we have determined all six of these numbers the triangle is solved.
It usually takes 3 of these six properties to solve a triangle.
combinations of 3 properties out of 6.
We will investigate the 8 possible
AAA
– There is not enough information to determine the length of the sides.
AAS ASA SAA
– Determine the third angle, then use the law of sines.
SSA ASS
– Use the law of sines to find a missing angle.
SAS
– Use the law of cosines to find the third side, and then either the law of sines or the law
of cosines to determine another angle.
SSS
– Use the law of cosines to find the angles.
Examples:
Example AAS:
Suppose that C = 25o, A = 55o;c = 10. Solve the triangle.
B = 180 o –(A+C) = 100 o
c/sinC = a/sinA = > a = c*sinA/sinC = 10* 0.819152/ 0.422618 = 19.38279
c/sinC = b/sinB = > b = c*sinB/sinC = 10* 0.984808/ 0.422618 = 23.30254
So the solution for this triangle is:
A = 55o
B = 100 o
C = 25o
a = 19.38279
b = 23.30254
c = 10
Example SSA with no solution.
a = 3,b = 20,A = 80 o
sinB/b = sinA/a = > sinB = b*sinA/a = 20* 0.984808/3 = 6.565385; However 6.565385 is larger than the
largest value the sine function can take so there is no solution.
Example SSA with two solutions.
a = 3, b = 20, A = 7 o
sinB = 20*sin7 o/3 = 20*0.121869/3 = 0.812462.
What angles have a sin value of .812462? 54.33721o and 125.6628o
So either B = 54.33721o or B = 125.6628o
If B = 54.33721o Then C = 180–(54.33721o+7 o) = 118.6628o
Thus, c = sinC*a/sinA = .877458*3/ 0.121869 = 21.59996
This solution is
A = 7o
B = 54.33721o
C = 118.6628o
a=3
b = 20
c = 21.59996
On the other hand, we could also have B = 125.6628o
If B = 125.6628o Then C = 180–(125.6628o +7 o) = 47.33721o
Thus, c = sinC*a/sinA = 0.735355*3/0.121869 = 18.10188
This solution is
A = 7o
B = 125.6628o
C = 47.33721o
a=3
b = 20
c = 18.10188
So that we have two complete solutions for the triangle.
Example SSA with exactly one solution
a = 10, b = 20, A = 60 o
sinB = 20*sin60 o/10 = 20* 0.866025/10 = 0.433013
What angles have a sin value of 0.433013? 25.65891o and 154.3411o
But 154.3411o+60o = 214.3411o > 180o, so only B = 25.65891o is usable.
C = 180–(A+B) = 180–(60+25.65891) = 94.34109o
c = a*sinC/sinA = 10* 0.997131/0.866025 = 11.51388
So the solution for this triangle is:
A = 60o
B = 25.65891o
C = 94.34109o
a = 10
b = 20
c = 11.51388
Example SAS:
b = 10; A = 50o,c = 20;
Then side a is given by
ܽଶ = ܾ ଶ + ܿ ଶ − 2ܾܿ cos ‫ܣ‬
a2 = 102+202–2*10*20*cos50o = 100+400–200cos50o
a2 = 242.885
a = 15.58477
sinB = b*sinA/a = 10*sin50o/15.58477 = 0.491534
Thus, B = arcSin(.491534) = 29.44146o
B could also be 180–29.44146o = 100.5585 o
So how do we decide which? Well, the larger angle is across from the larger side, and this side(b) is
smaller than side a, so pin an angle less than 50o
Alternatively, we could calculate B using the law of cosines:
cos ‫= ܤ‬
ܽଶ + ܿ ଶ − ܾ ଶ
2ܽܿ
cosB = (a2+c2–b2)/2ac = 15.584772+202–102)/(2*15.58477*20) = .87058
B = arcos(.87058) = 29.4416.
We have three ways to calculate C
C = 180–(A+B) = 180–79.4416 = 100.5585o
sinC = c*sinA/a = 20*sin50o/15.58477 = 0.983068
C = arcSin(0.983068) = 79.44146o
Or C = 180–79.44146o = 100.5585o
Finally, we could use the law of cosines:
cos ‫= ܥ‬
ܽଶ + ܾ ଶ − ܿ ଶ
2ܾܽ
cosC = (15.584772+102-202)/(2*15.58477*10) = −0.18324
C=arcos(−0.18324)=100.5585o
So the solution for this triangle is:
A = 50o
B = 29.44146o
C = 100.5585o
a = 15.58477
b = 10
c = 20
Example SSS:
a=10, b=15, c=20
We use the law of cosines:
cos ‫= ܥ‬
ܽଶ + ܾ ଶ − ܿ ଶ
2ܾܽ
cos ‫= ܤ‬
ܽଶ + ܿ ଶ − ܾ ଶ
2ܽܿ
cos ‫= ܣ‬
ܾ ଶ + ܿ ଶ − ܽଶ
2ܾܿ
cos C =
ଵ଴మ ାଵହమ ିଶ଴మ
ଶ∗ଵ଴∗ଵହ
= –0.25
cos B =
ଵ଴మ ାଶ଴మ ିଵହమ
ଶ∗ଵ଴∗ଶ଴
= 0.6875
cos A =
ଵହమ ାଶ଴మ ିଵ଴మ
ଶ∗ଵହ∗ଶ଴
= 0.875
C = arcCos( –0.25) = 104.4775o
B = arcCos(0.6875) = 46.56746o
A = arcCos(0.875) = 28.95502o
So the solution for this triangle is:
A = 28.95502o
B = 46.56746o
C = 104.4775o
a = 10
b = 15
c = 20