3_D_Trig

[Scremin34Egl]

Thanks you explain really well. I have another problem:

I thought of using the cosine rule but got stuck:
a) BD^2=x^2+x^2-2(x)(x)cos teata

 

[srmichael]

Thanks you explain really well. I have another problem:

I thought of using the cosine rule but got stuck:
a) BD^2=x^2+x^2-2(x)(x)cos teata

FIRST, please post one problem per thread.

SECOND, what you did is correct. You just need to put the finishing touches on it:

BD² = x² + x² - 2x²cosΘ

BD² = 2x² - 2x²cosΘ

BD² = 2x²(1 - cosΘ)

Now, repost in another thread and try parts (b) and (c) and let us know where you get stuck.

 

[Scremin34Egl]

FIRST, please post one problem per thread.

SECOND, what you did is correct. You just need to put the finishing touches on it:

BD² = x² + x² - 2x²cosΘ

BD² = 2x² - 2x²cosΘ

BD² = 2x²(1 - cosΘ)

Now, repost in another thread and try parts (b) and (c) and let us know where you get stuck.

Will do, but first how did you get 1-cos theta?

 

[srmichael]

Will do, but first how did you get 1-cos theta?

I factored out the 2x² from each term, leaving 1 - cosΘ.

 

[Scremin34Egl]

Okay makes sense, Starting with b), should I use cosine rule again because I see it has a square root on the other side and I have 2 sides and an included angle?

 

[Scremin34Egl]

Sorry, I think I should use the trig ratio tan because I see a tan in the denominator, right?

 

[srmichael]

Okay makes sense, Starting with b), should I use cosine rule again because I see it has a square root on the other side and I have 2 sides and an included angle?

Let me ask you, what does \(\displaystyle \tan\alpha\) equal? If you find \(\displaystyle \tan\alpha\) then I think you may see how to proceed to solve for h.

 

[srmichael]

Sorry, I think I should use the trig ratio tan because I see a tan in the denominator, right?

Correctamundo!

 

[Scremin34Egl]

Thanks man, I have solved b) and c). You have been of great help to me