Find Special Angles of a Triangle...1

Section 4.3

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

 

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.

 

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

 

Ok.

 

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

 

yes, I did radians and you do degrees

 

This is not the answer in the book.

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

 

This is what Ron Larson is looking for.

 

these are only radians

 

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

 

yes, 45°

 

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