Find Angles in a Triangle

[nycmathguy]

Section 6.4

Please do (44) in step by step fashion as a guide for me to do the rest.

 

[Country Boy]

Use the "cosine law": If the sides of a triangle are a, b, and c and the angles opposite each are A, B, and C then
a^2= b^2+ c^2- 2bc cos(A)
b^2= a^2+ c^2- 2ac cos(B) and
c^2= a^2+ b^2- 2ab cos(C)

so that cos(A)= (b^2+ c^2- a^2)/2bc etc.

In problem 43 the vertices are A= (1, 2), B= (3, 4), and C= (2, 5).
So a= sqrt((3- 2)^2+ (4- 5)^2)= sqrt(2)
b= sqrt((1- 2)^2+ (2- 5)^2)= sqrt(10) and
c= sqrt((1- 3)^2+ (2- 4)^2)= sqrt(8).

angle A is given by cos(A)= (10+ 8- 2)/2 sqrt(80)= 16/8 sqrt(5)
angle B is given by cos(B)= (2+ 8- 10)/2 sqrt(16)= 0 (a right angle! this is a right triangle!)
angle C is given by cos(C)= (2+ 10- 8)/2 sqrt(20)= 4/4 sqrt(5).

 

[nycmathguy]

Use the "cosine law": If the sides of a triangle are a, b, and c and the angles opposite each are A, B, and C then
a^2= b^2+ c^2- 2bc cos(A)
b^2= a^2+ c^2- 2ac cos(B) and
c^2= a^2+ b^2- 2ab cos(C)

so that cos(A)= (b^2+ c^2- a^2)/2bc etc.

In problem 43 the vertices are A= (1, 2), B= (3, 4), and C= (2, 5).
So a= sqrt((3- 2)^2+ (4- 5)^2)= sqrt(2)
b= sqrt((1- 2)^2+ (2- 5)^2)= sqrt(10) and
c= sqrt((1- 3)^2+ (2- 4)^2)= sqrt(8).

angle A is given by cos(A)= (10+ 8- 2)/2 sqrt(80)= 16/8 sqrt(5)
angle B is given by cos(B)= (2+ 8- 10)/2 sqrt(16)= 0 (a right angle! this is a right triangle!)
angle C is given by cos(C)= (2+ 10- 8)/2 sqrt(20)= 4/4 sqrt(5).

I will use your reply to help me solve the rest on my days off.

 

[nycmathguy]

Use the "cosine law": If the sides of a triangle are a, b, and c and the angles opposite each are A, B, and C then
a^2= b^2+ c^2- 2bc cos(A)
b^2= a^2+ c^2- 2ac cos(B) and
c^2= a^2+ b^2- 2ab cos(C)

so that cos(A)= (b^2+ c^2- a^2)/2bc etc.

In problem 43 the vertices are A= (1, 2), B= (3, 4), and C= (2, 5).
So a= sqrt((3- 2)^2+ (4- 5)^2)= sqrt(2)
b= sqrt((1- 2)^2+ (2- 5)^2)= sqrt(10) and
c= sqrt((1- 3)^2+ (2- 4)^2)= sqrt(8).

angle A is given by cos(A)= (10+ 8- 2)/2 sqrt(80)= 16/8 sqrt(5)
angle B is given by cos(B)= (2+ 8- 10)/2 sqrt(16)= 0 (a right angle! this is a right triangle!)
angle C is given by cos(C)= (2+ 10- 8)/2 sqrt(20)= 4/4 sqrt(5).

The question tells me to use vectors not the law of cosines. How is this done using vectors? Section 6.4 involves vectors.

 

[Country Boy]

The dot product of two vectors is the product of the lengths of the two vectors multiplied by the cos of the angle between them.

 

[nycmathguy]

The dot product of two vectors is the product of the lengths of the two vectors multiplied by the cos of the angle between them.

Can you show me how this is done as a guide for me to solve similar problems?

 

[MathLover1]

44.
the vertices are A= (-3, -4), B= (1, 7), and C= (8, 2)

to use vectors use same starting point for two vectors
for sides AB and AC as a vectors, starting point is A

so,

vector AB=<1-(-3), 7-(-4)> = <1+3, 7+4> =<4, 11>
vector AC= <8-(-3), 2-(-4)> = <8+3, 2+4> =<11, 6>

ange A is between vectors AB and AC

cos(A)=(AB*AC)/(||AB||*||AC||)

AB*AC=<4,11> *<11,6>=4*11+11*6=110
||AB||=sqrt(4^2+11^2)=sqrt(137)
||AC||=sqrt(11^2+6^2)=sqrt(157)

cos(A)=110/(sqrt(137)*sqrt(157))

cos(A)=110/sqrt(21509)

A=cos^-1(110/sqrt(21509))

A=0.722678 (result in radians)

< A=41.41°

same way you find angle B using vectors BA and BC, starting point is B
vector BA= <-3-1, -4-7>=<-4,-11>
vector BC=<8-1,2-7>=<7,-5>

BA*BC=(-4)*7+(-11*(-5))=27
||BA||=sqrt((-4)^2+(-11)^2)=sqrt(137)
||BC||=sqrt(7^2+(-5)^2)=sqrt(74)

cos(B)=27/(sqrt(137)*sqrt(74))
cos(B)=27/sqrt(10138)
B=cos^-1(27/sqrt(10138))
B=1.29931784439598213 (result in radians)
< B=74.45°

then angle C=180-(41.41°+74.45°)
< C=64.14°

 

[nycmathguy]

44.
the vertices are A= (-3, -4), B= (1, 7), and C= (8, 2)

to use vectors use same starting point for two vectors
for sides AB and AC as a vectors, starting point is A

so,

vector AB=<1-(-3), 7-(-4)> = <1+3, 7+4> =<4, 11>
vector AC= <8-(-3), 2-(-4)> = <8+3, 2+4> =<11, 6>

ange A is between vectors AB and AC

cos(A)=(AB*AC)/(||AB||*||AC||)

AB*AC=<4,11> *<11,6>=4*11+11*6=110
||AB||=sqrt(4^2+11^2)=sqrt(137)
||AC||=sqrt(11^2+6^2)=sqrt(157)

cos(A)=110/(sqrt(137)*sqrt(157))

cos(A)=110/sqrt(21509)

A=cos^-1(110/sqrt(21509))

A=0.722678 (result in radians)

< A=41.41°

same way you find angle B using vectors BA and BC, starting point is B
vector BA= <-3-1, -4-7>=<-4,-11>
vector BC=<8-1,2-7>=<7,-5>

BA*BC=(-4)*7+(-11*(-5))=27
||BA||=sqrt((-4)^2+(-11)^2)=sqrt(137)
||BC||=sqrt(7^2+(-5)^2)=sqrt(74)

cos(B)=27/(sqrt(137)*sqrt(74))
cos(B)=27/sqrt(10138)
B=cos^-1(27/sqrt(10138))
B=1.29931784439598213 (result in radians)
< B=74.45°

then angle C=180-(41.41°+74.45°)
< C=64.14°

This is very lengthy. I'll just take your word that this is the vector method to find the interior angles of the triangle.