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Trig Question

https://www.quora.com/profile/Bravewarrior/p-144933495
Here is the question and its solution. How is the gradient of the line basically tan theta?
Reply 1
Original post by pigeonwarrior
https://www.quora.com/profile/Bravewarrior/p-144933495
Here is the question and its solution. How is the gradient of the line basically tan theta?

Since the tan function is Opposite/Adjacent, we can see that we can create a triangle with y=3/4x, the x axis, and another side connecting the others (opposite the angle theta). The opposite (the vertical line, and therefore the y value) will be 3/4 of the x value (the adjacent, x-axis) since that is what the line's equation is. So tan theta is 3/4, the ratio between the side opposite the angle and the adjacent side. Hope this helped.
Reply 2
gradient = up/across

tan = up/across

:qed:
Reply 3
Original post by SigmaBoy69
Since the tan function is Opposite/Adjacent, we can see that we can create a triangle with y=3/4x, the x axis, and another side connecting the others (opposite the angle theta). The opposite (the vertical line, and therefore the y value) will be 3/4 of the x value (the adjacent, x-axis) since that is what the line's equation is. So tan theta is 3/4, the ratio between the side opposite the angle and the adjacent side. Hope this helped.

Thank you so much, it makes more sense now!
Reply 4
Original post by the bear
gradient = up/across

tan = up/across

:qed:

Thank you! 😁

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