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avatar+211 

Hi There,

 

I was hoping someone could show me the working to the below.

I feel i'm missing something quite simple.

 

 Aug 21, 2018
 #1
avatar+100803 
+3

Hi Vest4R,

 

\(3Tcos\theta - 1.5Tsin\theta = 0\\ T(3cos\theta - 1.5sin\theta )= 0\\ T=0\;\;or\;\; 3cos\theta - 1.5sin\theta=0\\ T=0\;\;or\;\; 3cos\theta = 1.5sin\theta\\ T=0\;\;or\;\; \frac{3cos\theta}{1.5cos\theta} = \frac{1.5sin\theta}{1.5cos\theta }\\ T=0\;\;or\;\; 2 = tan\theta\\ T=0\;\;or\;\; tan\theta=2\\ T=0\;\;or\;\; \theta=atan(2)\qquad\;\;NOTE: \theta\text{ must be in the first or third quad}\\ T=0\;\;or\;\; \theta=(63.43+180n)^\circ \qquad where \qquad n\in Z \)

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 Aug 21, 2018
 #2
avatar+211 
+2

Thanks Melody,

 

Can I ask where: \(\frac{3cos\theta}{1.5cos\theta} = \frac{1.5sin\theta}{1.5cos\theta}\) came from?

is this just dividing the left hand side by only half of its self?

 Aug 21, 2018
 #3
avatar+211 
+2

Actually, I just realized how silly I've been for the past couple of hours trying to work this out.....

 

\(3Tcos\theta - 1.5Tsin\theta = 0 \\ 3Tcos\theta =1.5Tsin\theta \\ 3T = 1.5T tan\theta \\ \frac{3}{1.5} = tan\theta \)

 

I appreciate your help for waking me up!!

 Aug 21, 2018

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