Trig question

[eco&math=die]

according to the question, I get costheta=c^2+b^2-a^2/2bc. However, I need to convert 'c' to a,b form, which I'm not very sure how to do it. Please help, thanks, guys.

 

[Dr.Peterson]

View attachment 30809
according to the question, I get costheta=c^2+b^2-a^2/2bc. However, I need to convert 'c' to a,b form, which I'm not very sure how to do it. Please help, thanks, guys.

You just used the Law of Cosines, and didn't use angle ABC.

Try the Law of Sines.

When you type equations, be careful to use parentheses as needed. You meant "cos theta=(c^2+b^2-a^2)/(2bc)". What you wrote means [imath]\cos\theta=c^2+b^2-\frac{a^2}{2bc}[/imath], or possibly even [imath]\cos\theta=c^2+b^2-\frac{a^2}{2}bc[/imath].

 

[lev888]

View attachment 30809
according to the question, I get costheta=c^2+b^2-a^2/2bc. However, I need to convert 'c' to a,b form, which I'm not very sure how to do it. Please help, thanks, guys.

You don't need c. Use the sin of double angle formula and the law of sines.

 

[Deleted member 4993]

View attachment 30809
according to the question, I get costheta=c^2+b^2-a^2/2bc. However, I need to convert 'c' to a,b form, which I'm not very sure how to do it. Please help, thanks, guys.

cos(Θ) =(c^2 + b^2-a^2) / (2*b*c)

Those parentheses are super important.

Use the fact that angle BCA = 180^o - 3Θ & laws of Sines.

 

[eco&math=die]

You just used the Law of Cosines, and didn't use angle ABC.

Try the Law of Sines.

When you type equations, be careful to use parentheses as needed. You meant "cos theta=(c^2+b^2-a^2)/(2bc)". What you wrote means [imath]\cos\theta=c^2+b^2-\frac{a^2}{2bc}[/imath], or possibly even [imath]\cos\theta=c^2+b^2-\frac{a^2}{2}bc[/ [/QUOTE][/imath]

I used the angle ABC and get sin 2Θ=2 sinΘ cosΘ, then use the sines rule, which (2 sinΘ cosΘ)/b=(sinΘ)/a, I get cosΘ=b/(2a). And it is the correct answer. Thank you so much!!! Also, I will be very careful about my parentheses from now on.

 

[eco&math=die]

You don't need c. Use the sin of double angle formula and the law of sines.

Thank you so much, this is very helpful!!!

 

[eco&math=die]

cos(Θ) =(c^2 + b^2-a^2) / (2*b*c)

Those parentheses are super important.

Use the fact that angle BCA = 180^o - 3Θ & laws of Sines.

Thank you for helping me! It is super helpful!