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How do i resolve this? [M1 Edexcel]


Where should the alpha angle be for Tb?

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Reply 1
Bump!! :smile:
since Ta and Tb have perpendicular directions and the vertical direction is perpendicular to AB (the horizontal direction), the angle you are looking for, is between the Tb and the vertical direction.
(Two acute angles with perpendicular sides are equal )
(edited 7 years ago)
Reply 3
Original post by depymak
since Ta and Tb have perpendicular directions and the vertical direction is perpendicular to AB (the horizontal direction), the angle you are looking for, is between the Tb and the vertical direction.
(Two acute angles with perpendicular sides are equal )


I still don't get it? :s-smilie:
Reply 4
Bump!!

Where should the alpha angle be for Tb?
Original post by Adorable98

Where should the alpha angle be for Tb?


w1w1w.png
Reply 6
An alternate method is also drawing a triangle of forces. The two cables are perpendicular, giving you a right angled triangle with 1000g as the hypotenuse. You can then use trig to find tension in A, then pythag to find tension in B.


Posted from TSR Mobile
(edited 7 years ago)
Original post by EricPiphany
w1w1w.png


Why is the angle (90 - a)?
Original post by Adorable98
Bump!!

Where should the alpha angle be for Tb?


Just a hint but you know that

tanθ=oa=34tan\theta = \frac{o}{a} = \frac{3}{4}

What does this tell you about the triangle and the angles in the triangle?
Original post by Don Pedro K.
Why is the angle (90 - a)?


tan(alpha) = 3/4, it follows directly from this, you can spot something about the triangle
Original post by Adorable98
I still don't get it? :s-smilie:

Sorry,
what about this sketch?
https://drive.google.com/file/d/0B_iqJImnYHZ6Yi1IRXVNRE9JU2c/view?usp=sharing
Reply 11
angle_trouble.png
See diagram.
Original post by EricPiphany
w1w1w.png


What's the name of the 'rule' when finding α for tb:smile:
cos for Ta I know the angle is alpha because it is alternate to the α on parallel lines (Z shape)
Original post by natninja
Just a hint but you know that

tanθ=oa=34tan\theta = \frac{o}{a} = \frac{3}{4}

What does this tell you about the triangle and the angles in the triangle?

I really don't know :redface:
Original post by Adorable98
What's the name of the 'rule' when finding α for tb:smile:
cos for Ta I know the angle is alpha because it is alternate to the α on parallel lines (Z shape)

I really don't know :redface:


Learn this, it will help if a similar thing comes up - if tan(angle) = 3/4 then it is a right angled triangle as the other side must have length 5 (so you also know that sin(angle) = 3/5 and cos(angle) = 4/5)
Original post by Adorable98
What's the name of the 'rule' when finding α for tb:smile:
cos for Ta I know the angle is alpha because it is alternate to the α on parallel lines (Z shape)

I really don't know :redface:


w1w1w.png

What is β\beta. Notice it's common to two right angles, so their remainders are equal.
Original post by EricPiphany
w1w1w.png

What is β\beta. Notice it's common to two right angles, so their remainders are equal.


Original post by natninja
Learn this, it will help if a similar thing comes up - if tan(angle) = 3/4 then it is a right angled triangle as the other side must have length 5 (so you also know that sin(angle) = 3/5 and cos(angle) = 4/5)


Original post by OscarTG
angle_trouble.png
See diagram.




Original post by SkyJP
An alternate method is also drawing a triangle of forces. The two cables are perpendicular, giving you a right angled triangle with 1000g as the hypotenuse. You can then use trig to find tension in A, then pythag to find tension in B.


Posted from TSR Mobile


Thank you!! :h::h:
Are the triangles considered to be similar triangles? :redface:
holy crap I just went back to the question and saw the right angle between the two cables...LOOOOOOOOOOOOL that makes things so much easier hahahaha I'm such an idiot.
Original post by Don Pedro K.
holy crap I just went back to the question and saw the right angle between the two cables...LOOOOOOOOOOOOL that makes things so much easier hahahaha I'm such an idiot.


I somehow managed to get the trig identity sin^x + cos^x = 1 in there to get the answers >.> lololl
i get the question but im not getting the answers, anyone got some solutions?

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