Navigation...Find Bearings of Last Two Legs

[nycmathguy]

Section 6.2

Can you do 51 for me?

 

[MathLover1]

51.

The bearing of 2 last legs of race are angels A and C

Applying the Law of cosine we have:

cos(C)= (a² + b² -c²)/(2ab)

cos(C)= (1700² + 3000² -3700²)/(2*1700*3000)

cos(C)=(11890000-13690000)/10200000

cos(C)=-3/17

C=cos^-1(-3/17)

C=100.2°


cos(A) = (b^2 + c^2 - a^2)/(2bc)

cos(A) = (3000^2 +3700^2 - 1700^2)/(2*3000*3700)

cos(A) = 33/37

A=cos^-1(33/37)

A=26.89°

The last two bearings are 26.89° and 100.2°

 

[nycmathguy]

51.

View attachment 1704



The bearing of 2 last legs of race are angels A and C

Applying the Law of cosine we have:

cos(C)= (a² + b² -c²)/(2ab)

cos(C)= (1700² + 3000² -3700²)/(2*1700*3000)

cos(C)=(11890000-13690000)/10200000

cos(C)=-3/17

C=cos^-1(-3/17)

C=100.2°


cos(A) = (b^2 + c^2 - a^2)/(2bc)

cos(A) = (3000^2 +3700^2 - 1700^2)/(2*3000*3700)

cos(A) = 33/37

A=cos^-1(33/37)

A=26.89°

The last two bearings are 26.89° and 100.2°

You said:

"The bearing of 2 last legs of race are angels A and C."

I know you meant to say angles not angels.

What in the question indicates this fact about angles A and C?

 

[MathLover1]

race starts at B, continues to C (bearing first time), continues to A (bearing second time), continues back to B

 

[nycmathguy]

race starts at B, continues to C (bearing first time), continues to A (bearing second time), continues back to B

Ok. Thank you.