Specialist 1/2 Question Thread!

[ErnieTheBirdi]

Show that the sum of the first 2n terms of an arithmetic sequence is n times the sum of the two middle terms.

Can someone please explain this question to me please, I don’t understand it even after looking at the worked solutions?

[MoonChild1234]

If sin (x)=0.3, cos (a) =0.5, tan (b) =2.4, and x, a and b are in the first quadrant. Find the values of the following:

 sin (pi-a)

I got 1/2 but the answers say root 3/2

i just wanted some assistance on how to get to that answer.
 thanks in advance!

[MoonChild1234]

Evaluate (sin a + cos a)^2 + (sin a - cos a)^2

I also wasn't really sure about this one, thanks again!

[^^^111^^^]

If sin (x)=0.3, cos (a) =0.5, tan (b) =2.4, and x, a and b are in the first quadrant. Find the values of the following:

 sin (pi-a)

I got 1/2 but the answers say root 3/2

i just wanted some assistance on how to get to that answer.
 thanks in advance!

(a) = cos^-1(0.5)
sin(cos^-1(0.5)) = squareroot(3)/2
Using the identity that sin(π-θ)=sinθ, we get that sin(π-a) is equal to sin (a) which is equal to squareroot(3)/2.

Evaluate (sin a + cos a)^2 + (sin a - cos a)^2

I also wasn't really sure about this one, thanks again!

Using the formulas (x+y)^2=x²+2xy+y² and  (x-y)^2=x²-2xy+y²,
we get that sin²a+cos²a+2*sin(a)(cos)(a) and sin²a+cos²a-2*sin(a)(cos)(a).
Adding those two expressions, we get 2(sin²(a)) + 2(cos²(a)) = 2(sin²(a))+(cos²(a)).
Using the identity sin²(a)+cos²(a)=1, we get 2(1)=2. Therefore, your answer should be 2.

[MoonChild1234]

hi! i was wondering how to do these questions:

|x-4|-|x-2|=6

|2x-5|-|4-x|=10

[S_R_K]

hi! i was wondering how to do these questions:

|x-4|-|x-2|=6

|2x-5|-|4-x|=10

Use the definition of |x| to break into cases and solve, making sure to check your solutions against the domain. eg:

If x > 4, then x – 4 - (x – 2) = 6, gives no solution.
If 4 > x > 2, then –(x – 4) – (x – 2) = 6 gives x = 0. This solution is rejected because it is outside the domain.
If 2 > x, then –(x – 4) – (–(x – 2)) = 6 gives no solution.

Hence there is no solution to |x – 4| – |x – 2| = 6.

[MoonChild1234]

thank you! but how would i know what cases to break it into?

[fun_jirachi]

thank you! but how would i know what cases to break it into?

In general, for some equation \(|x-a| + |x-b| = c \ (a < b)\), you're going to have three cases - \(x < a, a \leq x \leq b, x > b\). In most cases of equations in this form, you'll have what we call a bucket curve (try graphing one of your examples on Desmos or GeoGebra! it'll make sense ). Each of the three cases represents each 'part' of the bucket. Muck around a bit with the equations, and hopefully it'll start making a bit more sense as to where the cases come from

Hope this helps

[xpikachuzz]

https://imgur.com/etBnp6h . Find angle COD where O is the center and abcd lie on the circumference. context: my school is closing tomorrow and i had a sac today, got stumped on this question.
If it helps this topic was on congruence (of triangles), similarity (of triangles) and circle theorems.

[S_R_K]

https://imgur.com/etBnp6h . Find angle COD where O is the center and abcd lie on the circumference. context: my school is closing tomorrow and i had a sac today, got stumped on this question.
If it helps this topic was on congruence (of triangles), similarity (of triangles) and circle theorems.

∠BCA = 40° (half the angle at the centre subtended by the same arc).
∠APC = 128°
Hence, ∠CAD = 12°.
Hence ∠COD = 24° (using the same theorem as above).

[xpikachuzz]

∠BCA = 40° (half the angle at the centre subtended by the same arc).
∠APC = 128°
Hence, ∠CAD = 12°.
Hence ∠COD = 24° (using the same theorem as above).

Thank you i feel so dumb for not getting this.

[S_R_K]

Hi

I really need help with this question:

You just need to find the value of x

Thanks

A couple of bits of circle geometry that are useful: (1) The point where the tangents intersect and the centres of the two circles are collinear. (2) The line passing through these points bisects the angle between the tangents.

[SS1314]

Hey

I needed help with another question:

A tunnel 300 m long has a semicircular cross-section. It requires temporary supports, AB and BC, as shown.
Point B is vertically above D. If AB = 15 m and BC = 8 m, find:
(a)   the diameter of the tunnel

(b)   the height of B above the floor of the tunnel

(c)   the distance D is from the wall at C

(d)   the maximum height a 3-metre-wide truck could be and still pass under the supports.

I just needed help with question d


Thanks 

[fun_jirachi]

Hey there!

Here's a diagram of the situation we have in part (d):


Basically, the max height we have is if the truck passes right through the middle. We have that the tunnel is 17m wide, which is where all the numbers come from. Thus we can also deduce the radius - which I haven't labelled, then use Pythagoras' theorem to find the max height h as labelled on the diagram.

Hope this helps

[S_R_K]

Hey there!

Here's a diagram of the situation we have in part (d):
(Image removed from quote.)

Basically, the max height we have is if the truck passes right through the middle. We have that the tunnel is 17m wide, which is where all the numbers come from. Thus we can also deduce the radius - which I haven't labelled, then use Pythagoras' theorem to find the max height h as labelled on the diagram.

Hope this helps

You are assuming that the truck drives down the middle of the tunnel. The truck can be taller if it doesn't drive down the middle (because of the location of the supports AB and BC).