I understand how to prove it without using the triangle, but I clearly need to use it in some way, that is what is confusing me
I guess you're refering to the cos2θ+sin2θ=1 identity.
Looking at the triangle(s), we could use Pythagoras either on ABD or on ABC as our starting point, though it's only one step removed from the trig identity.
I guess you're refering to the cos2θ+sin2θ=1 identity.
Looking at the triangle(s), we could use Pythagoras either on ABD or on ABC as our starting point, though it's only one step removed from the trig identity.
thanks! that would work, but then why would they show both triangles?
Actually there are three similar triangles there. Everything relating to D is redundant if we use Pythagoras.
The only thing I can think of is they want you to avoid using Pythagoras, and work with the similar triangles - in which case they should have specified that. The exact way to do that eludes me at present though - assuming it is feasible.
Actually there are three similar triangles there. Everything relating to D is redundant if we use Pythagoras.
The only thing I can think of is they want you to avoid using Pythagoras, and work with the similar triangles - in which case they should have specified that. The exact way to do that eludes me at present though - assuming it is feasible.
Is the first line correct? You seem to have leg > hypotenuse? Also x isn't defined?
It should just be a couple of lines, one to write down Pythagoras on ABC, the other to replace the lengths with the equivalent trig terms. As noted above, I can't see how the triangle split makes it any simpler (can't get much simpler than a couple of lines) unless they want you to imagine ADB as the starting point, then consider adding BDC to it to get ABC. You could then note dividing by cos on ABD does the "same" as adding BDC to get ABC. You're constructing a new, similar triangle with the desired length property.
You could use similar triangles without assuming Pythagoras (noted above and basically proving Pythagoras) as AC = AD + DC and get AD and DC using similar triangles to ABC. A bit longer though and more complex.
Is the first line correct? You seem to have leg > hypotenuse? Also x isn't defined?
It should just be a couple of lines, one to write down Pythagoras on ABC, the other to replace the lengths with the equivalent trig terms. As noted above, I can't see how the triangle split makes it any simpler (can't get much simpler than a couple of lines) unless they want you to imagine ADB as the starting point, then consider adding BDC to it to get ABC. You could then note dividing by cos on ABD does the "same" as adding BDC to get ABC. You're constructing a new, similar triangle with the desired length property.
You could use similar triangles without assuming Pythagoras (noted above and basically proving Pythagoras) as AC = AD + DC and get AD and DC using similar triangles to ABC. A bit longer though and more complex.
I just realised that I used the identity to prove the solution lol, I didn't use the triangle However I did it again and found a solution lol wait a sec lemme post it
Is the first line correct? You seem to have leg > hypotenuse? Also x isn't defined?
It should just be a couple of lines, one to write down Pythagoras on ABC, the other to replace the lengths with the equivalent trig terms. As noted above, I can't see how the triangle split makes it any simpler (can't get much simpler than a couple of lines) unless they want you to imagine ADB as the starting point, then consider adding BDC to it to get ABC. You could then note dividing by cos on ABD does the "same" as adding BDC to get ABC. You're constructing a new, similar triangle with the desired length property.
You could use similar triangles without assuming Pythagoras (noted above and basically proving Pythagoras) as AC = AD + DC and get AD and DC using similar triangles to ABC. A bit longer though and more complex.
Looks ok, there are other solutions, some simpler. Don't know if you'd get full marks in an exam. There is little/no explanation of what you're doing or why. Even if it's just so the examiner marks it correctly or so you can check your own work, explanations help. Hard to comment too much on the cheating bit apart from to say your first post was "looking for solutions online" and you don't explain the working ... Maybe they have a point?
Looks ok, there are other solutions, some simpler. Don't know if you'd get full marks in an exam. There is little/no explanation of what you're doing or why. Even if it's just so the examiner marks it correctly or so you can check your own work, explanations help. Hard to comment too much on the cheating bit apart from to say your first post was "looking for solutions online" and you don't explain the working ... Maybe they have a point?
Well I never did find an answer to any of the questions so I could't have cheated. Also I'm pretty sure you don't have to explain it with words if you're able to explain it with maths
Looks ok, there are other solutions, some simpler. Don't know if you'd get full marks in an exam. There is little/no explanation of what you're doing or why. Even if it's just so the examiner marks it correctly or so you can check your own work, explanations help. Hard to comment too much on the cheating bit apart from to say your first post was "looking for solutions online" and you don't explain the working ... Maybe they have a point?
Well I never did find an answer to any of the questions so I could't have cheated. Also I'm pretty sure you don't have to explain it with words if you're able to explain it with maths