Find Special Angles of a Triangle...1

[nycmathguy]

Section 4.3

Can you do 53 (a & b) as a guide for me to do a few more?

 

[MathLover1]

53
a)

sin(theta)=1/2

1/2=opp/hyp=> opp=1, hyp=

adj=sqrt(2^2-1^2)=sqrt(3)

cos(theta)=sqrt(3)/2

tan(theta)=opp/adj=1/sqrt(3)

cot(theta)=1/tan(theta)=1/(1/sqrt(3))=sqrt(3)

sec(theta)=1/cos(theta)=1/(sqrt(3)/2)=2/sqrt(3)

csc(theta)=1/sin(theta)=1/(1/2)=2


b)

csc(theta)=2

csc(theta)=1/sin(theta)

1/sin(theta)=2

sin(theta)=1/2=> opp/hyp

opp=1, hyp=2

adj=sqrt(2^2-1^2)=sqrt(3)

cos(theta)=adj/hyp

cos(theta)=sqrt(3)/2


tan(theta)=opp/adj=1/sqrt(3)

cot(theta)=1/tan(theta)= 1/(1/sqrt(3))=sqrt(3)

sec(theta)=1/cos(theta)=1/(sqrt(3)/2)=2/sqrt(3)

csc(theta)=1/sin(theta)=1/(1/2)=2

 

[nycmathguy]

53
a)

sin(theta)=1/2

1/2=opp/hyp=> opp=1, hyp=

adj=sqrt(2^2-1^2)=sqrt(3)

cos(theta)=sqrt(3)/2

tan(theta)=opp/adj=1/sqrt(3)

cot(theta)=1/tan(theta)=1/(1/sqrt(3))=sqrt(3)

sec(theta)=1/cos(theta)=1/(sqrt(3)/2)=2/sqrt(3)

csc(theta)=1/sin(theta)=1/(1/2)=2


b)

csc(theta)=2

csc(theta)=1/sin(theta)

1/sin(theta)=2

sin(theta)=1/2=> opp/hyp

opp=1, hyp=2

adj=sqrt(2^2-1^2)=sqrt(3)

cos(theta)=adj/hyp

cos(theta)=sqrt(3)/2


tan(theta)=opp/adj=1/sqrt(3)

cot(theta)=1/tan(theta)= 1/(1/sqrt(3))=sqrt(3)

sec(theta)=1/cos(theta)=1/(sqrt(3)/2)=2/sqrt(3)

csc(theta)=1/sin(theta)=1/(1/2)=2

Thanks. I will try several tomorrow and post here.

 

[MathLover1]

try to use wolfram now and double check on one of the correct answers

 

[nycmathguy]

try to use wolfram now and double check on one of the correct answers

Ok.

 

[nycmathguy]

try to use wolfram now and double check on one of the correct answers

This problem tells us to find each value of theta in degrees and radians.

 

[MathLover1]

This problem tells us to find each value of theta in degrees and radians.

yes, I did radians and you do degrees

 

[nycmathguy]

53
a)

sin(theta)=1/2

1/2=opp/hyp=> opp=1, hyp=

adj=sqrt(2^2-1^2)=sqrt(3)

cos(theta)=sqrt(3)/2

tan(theta)=opp/adj=1/sqrt(3)

cot(theta)=1/tan(theta)=1/(1/sqrt(3))=sqrt(3)

sec(theta)=1/cos(theta)=1/(sqrt(3)/2)=2/sqrt(3)

csc(theta)=1/sin(theta)=1/(1/2)=2


b)

csc(theta)=2

csc(theta)=1/sin(theta)

1/sin(theta)=2

sin(theta)=1/2=> opp/hyp

opp=1, hyp=2

adj=sqrt(2^2-1^2)=sqrt(3)

cos(theta)=adj/hyp

cos(theta)=sqrt(3)/2


tan(theta)=opp/adj=1/sqrt(3)

cot(theta)=1/tan(theta)= 1/(1/sqrt(3))=sqrt(3)

sec(theta)=1/cos(theta)=1/(sqrt(3)/2)=2/sqrt(3)

csc(theta)=1/sin(theta)=1/(1/2)=2

This is not the answer in the book.

For 53 part (a), the answer is 30° = pi/6. Same for part (b).

 

[nycmathguy]

yes, I did radians and you do degrees

This is what Ron Larson is looking for.

 

[MathLover1]

This is what Ron Larson is looking for.

View attachment 919

View attachment 920

View attachment 921

these are only radians

 

[nycmathguy]

these are only radians

I also posted the values in degrees. For example, pi/4 = 45°. No?

 

[MathLover1]

yes, 45°

 

[nycmathguy]

yes, 45°

I listed my answers in both degrees and radians. Take a look again.