Q4. Flying Fox - Geometry.

[nickk]

Yer i used TAN to find the angle in the triangle, and tan again to find the height in the middle and added the height of the square at the bottom

[gamblor0]

LOL its 7.75.

im probably the only one in the state that solved it this way, but i worked out the EQUATION of the line and it was y=-0.5x+12.5.

then simply sub in 95 and wallah!!! 7.75

[Studyinghard]

LOL its 7.75.

im probably the only one in the state that solved it this way, but i worked out the EQUATION of the line and it was y=-0.5x+12.5.

then simply sub in 95 and wallah!!! 7.75

LOL its 7.75.

im probably the only one in the state that solved it this way, but i worked out the EQUATION of the line and it was y=-0.5x+12.5.

then simply sub in 95 and wallah!!! 7.75

wow, thats quite pro, gotta admit

[Xavier1234]

  :idiot2:

LOL its 7.75.

im probably the only one in the state that solved it this way, but i worked out the EQUATION of the line and it was y=-0.5x+12.5.

then simply sub in 95 and wallah!!! 7.75

how did you deduce the equation of the line??? and using that actually comes up as -35.  :buck2:

[cjudd3votes]

LOL its 7.75.

im probably the only one in the state that solved it this way, but i worked out the EQUATION of the line and it was y=-0.5x+12.5.

then simply sub in 95 and wallah!!! 7.75

Yeah, used the same method, but you could've used stacks of different ways.

[travy92]

:idiot2:

LOL its 7.75.

im probably the only one in the state that solved it this way, but i worked out the EQUATION of the line and it was y=-0.5x+12.5.

then simply sub in 95 and wallah!!! 7.75

how did you deduce the equation of the line??? and using that actually comes up as -35.  :buck2:

Used the same method, lol but as I said somewhere else. We'll probably lose a mark since we were SUPPOSED to use similar triangles.. (Geometry section lol, not graphs/relations haha).

Anyway to actually achieve this, let T = 0.
So you have the very left side where the flying fox is located at the top of the hill, and since that's x=0, the y value is the height.
So: (0,12.5)
Then do the same with the other end. (160,4.5) since 160m from T and 4.5m high.
Now you have two points:
(0,12.5)
(160,4.5)
Use m=y2-y1/x2-x1
And you get -0.05.
Then y-y1=m(x-x1)
to get C value which turns out to be: c=12.5
So the equation you're left with: y=-0.05x + 12.5
Sub in value of H (95m)
And you get: 7.75m

Done.

[cjudd3votes]

:idiot2:

LOL its 7.75.

im probably the only one in the state that solved it this way, but i worked out the EQUATION of the line and it was y=-0.5x+12.5.

then simply sub in 95 and wallah!!! 7.75

how did you deduce the equation of the line??? and using that actually comes up as -35.  :buck2:

Used the same method, lol but as I said somewhere else. We'll probably lose a mark since we were SUPPOSED to use similar triangles.. (Geometry section lol, not graphs/relations haha).

Anyway to actually achieve this, let T = 0.
So you have the very left side where the flying fox is located at the top of the hill, and since that's x=0, the y value is the height.
So: (0,12.5)
Then do the same with the other end. (160,4.5) since 160m from T and 4.5m high.
Now you have two points:
(0,12.5)
(160,4.5)
Use m=y2-y1/x2-x1
And you get -0.05.
Then y-y1=m(x-x1)
to get C value which turns out to be: c=12.5
So the equation you're left with: y=-0.05x + 12.5
Sub in value of H (95m)
And you get: 7.75m

Done.

Didn't specify which method and it's Further, not methods. I'd say 7.75 with any working would get 2 marks