Random quiiestions from maths assignments help! (1 Viewer)

[jackerino]

Im stuck on a few questions of my maths assignment, quite a few actually. Please help someone!!??!

Q2. Simplify (n-1)C2+nC2 (Its a combination thing). State any restrictions that must be placed on n

Q3. (Idk how to go solve the resulting cubic) Solve 15/(2x-1)>or equal to x

Q7. Heres how to draw it: Draw a scalene triangle, and label the bottom right angle A, and clockwise add B and C to other angles. Now on BA there is a midpoint T. Connect T to C. ALso, on BC there is a point P. Connect A to P. Now AP intersects TC at E. AE is perpendicular to CT. TC bisects angle C. Connect E with M.
i) Prove triangle ACE is congruent to triangle PCE ( I cant find a third reason, only two)
ii) Explain why AE= EP (Matching sides congruent triangles from i)
iii) Hence prove EM is parallel to PB
This question is damn easy lolz I just cant seem to get the right result, how stupid of me -_-


Q12. A point P(x,y) moves so that the sum of the squares of its distance from the points A(0,4), B(0,-4), and C(6,3) is 77. Show that the locus is a circle and state its centre and radius.


Q15. Here's how to draw diagram: Draw a right angled triangle ABC, labelled clockwise from A at the top. Make C face west of the page. Now with AB as a common side, draw a reflection of the triangle, except extend the hypotenuse and AB further so A reaches to D, which is past B, and the hypotenuse reaches to E, such that the formed angle of ADE =90 degrees. Now connect C to E. CE intersects BD at F. Overall, triangle ABC is similar to triangle AD, and AB=a units. The ratio of similarity of AB:AD is k:l

Here are the questions:
a) Prove triangle FBC is similar to triangle FDE
b) find the ratio of the lengths BF to FD
c) SHow that FD=(al(l-k))/(k(l+k))
d) Given AB:AD=2:3, and FD is an integer, find the possible values of A.


Q16.

a) Prove tan(a+b+c)=(tana+tanb+tanc-tan(a)tan(b)tan(c))/(1-tanatanb-tanatanc-tanbtanc)
i)If a, b and c are the angles of a triangle, explain why tana+tanb+tanc-tanatanbtanc=0
b)Given tana/2=tanba/3=tanc/4=k, show that k=sqrt(6)/4
c) Hence find the size of a in degrees and minutes




SORRY ITS SO BIG! Its just super hard and im a bit confused. PLease help with working out!

 

[jackerino]

If someone could help with at least one question It'd be great

 

[Carrotsticks]

2) Turn the nCk notation to factorial notation, then cross multiply and simplify.

3) Guess one integer solution and once you have one, factorise your cubic to get a linear expression multiplied with a quadratic. Then factorise the quadratic.

7) Compute |PA|^2 + |PB|^2 + |PC|^2 = 77 and complete the square in terms of x and y to get the circle.

16) Consider the expansion of tan(A+B+C) using compound angle formula, but note that since A+B+C=180 (angle sum triangle), then tan(A+B+C)=0. So let the expansion be zero, and you will get the answer.

 

[jackerino]

Thanks carrotsticks! Okay well there's one more which I forgot to add, "Describe two ways of solving the equation sintheta=tan(theta/2), and solve using one of those ways.

 

[theind1996]

Thanks carrotsticks! Okay well there's one more which I forgot to add, "Describe two ways of solving the equation sintheta=tan(theta/2), and solve using one of those ways.

t-formulae and auxiliary angle?

 

[Carrotsticks]

1) Use the 't formula' method, where t = tan(x/2) and so sin(x) = 2t/(1+t^2)

2) Split sin(x) into 2sin(x/2)cos(x/2) and the tan(x/2) into sin(x/2)/cos(x/2), then re-arrange and solve in terms of x/2.

 

[Carrotsticks]

t-formulae and auxiliary angle?

Can't use auxiliary angles here because the angle is different and we have the tangent function (can be decomposed into sin/cos but still doesn't work)

 

[theind1996]

Can't use auxiliary angles here because the angle is different and we have the tangent function (can be decomposed into sin/cos but still doesn't work)

oh yea, shit lol.

 

[jackerino]

Okay found another thing im unsure of,
"The point M(x,y) divides the interval joining P(x1,y1) and Q(x2,y2) into the ratio k:l"
Ive done a), which says to show by geometry that k/l=(y-y1)/(y2-y), then in part b) it says "Hence show that the equation of the line PQ is
(x-x1)(y-y2)=(y-y1)(x-x2)

Idk how to use part a) for part b), so I found the equation of the line manually, which ended up as
(x-x1)(y2-y1)=(x2-x1)(y-y1)

So I think I will be marked wrong, cos obviously they said to find THAT particular way, and I found another way, so how DO you use part a) to get part b)?

 

[jackerino]

2) Turn the nCk notation to factorial notation, then cross multiply and simplify.

3) Guess one integer solution and once you have one, factorise your cubic to get a linear expression multiplied with a quadratic. Then factorise the quadratic.

7) Compute |PA|^2 + |PB|^2 + |PC|^2 = 77 and complete the square in terms of x and y to get the circle.

16) Consider the expansion of tan(A+B+C) using compound angle formula, but note that since A+B+C=180 (angle sum triangle), then tan(A+B+C)=0. So let the expansion be zero, and you will get the answer.

Oh with the last one. 16), idk how to expand tan(a+b+c)? and for 7), thats the DISTANCE PA^2+PB^2...etc right?

 

[jackerino]

HEY GUYS if you have a small scalene triangle embedded in a larger scalene triangle, like this:
http://www.ricksmath.com/images/triangle2.jpg
but with a point P on AB connected to a point M on CB,

how do you prove PM is parallel to AC? AP:AB=1:2, and CM:CB is 1:2.

 

[jackerino]

Please someone help me! :'(

 

[Carrotsticks]

Oh with the last one. 16), idk how to expand tan(a+b+c)? and for 7), thats the DISTANCE PA^2+PB^2...etc right?

To expand tan(a+b+c), you can bracket it as tan((a+b)+c) then use compound angle formula, then expand the tan(a+b) again.

And for 7, yep that's the distance.

HEY GUYS if you have a small scalene triangle embedded in a larger scalene triangle, like this:
http://www.ricksmath.com/images/triangle2.jpg
but with a point P on AB connected to a point M on CB,

how do you prove PM is parallel to AC? AP:AB=1:2, and CM:CB is 1:2.

Not quite sure what you mean here.

 

[jackerino]

Not quite sure what you mean here.[/QUOTE]


Erm, basically its a scalene triangle within another scalene triangle, such that the smaller one is similar to the larger one. The similarity is easily proven by the two sets of corresponding angles, though I have to prove that the two lines that cross the two short bits of each triangle are parallel. THe two parallel lines!>?! oh dear idk how to explain... Does that make sense?

And thanks for Q7. solved with your help