[SOLVED] Little Help with Math Question

[Suhail]

Right I am practising a few last things for my Maths Exam on Monday, I've been doing loads of past papers but there is this just 1 question which I want to make sure I got correct (Its a A* question worth quite a bit of marks). So any help or input would be appreciated . Here is the question:

 

 

ABC is a Right Angled Triangle

P is a point on AB such that AP = 4PB

BC = 3PB

Angle ACB = 90 Degrees

 

Show that sin B = k / 5

 

Write the Value of k where k is a positive integer

 

Sorry the drawing is bad, did a quick photoshop job to represent the image in the exam paper

Right now above is the information that is given. Here is what I have achieved so far. (Please note that this is on the Non-Calculator Paper ):

 

1. Label what angle we want:

 

2. Write in the algebra using the information given. (x represents the value of PB):

 

3. Since the Sine rule is Opposite over Hypotenuse we need to label them in to our triangle:

 

4. We need to find the value for the Opposite side. Time to do some Pythagoras:

 

5. So now I divide Opposite over Hypotenuse:

 

6. Right now we can consider that the x's cancel each other out becuase whatever number you put when you simplify the fractions you get:

 

Thats what I had so far. My question is that is the answer k = 4? And if so how would you explain the last step becuase you can't exactly say that they cancel each other out. Or can you?

 

Thanks

- Suhail

[derekreid]

Yup, looks right. If you solve for your angle B (using inverse sin), you get 53 degrees, which is reasonable.

 

You can cancel the x's in the last step.

 

Good luck Monday.

[Numberzz]

sin=opp/hyp

cos=adj/hyp

tan=opp/adj

 

Using the inverse sin function, where B=4/5, the angle comes out to be 53.13º.

 

I'll explain the rest:

Hope that helps.

[skyhighmac]

Good job Numberzz. A way to remember the sin/cos/tan stuff is this: Soh Cah Toa (Pronounced So Ka Toe-Ha)

The letters stand for:

S > Sine

O > Oppisiter

H > Hypotenuse

A > Adjacent

[Numberzz]

Good job Numberzz. A way to remember the sin/cos/tan stuff is this: Soh Cah Toa (Pronounced So Ka Toe-Ha)

The letters stand for:

S > Sine

O > Oppisiter

H > Hypotenuse

A > Adjacent

No, that's a dumb way to remember. The correct way is:

[Suhail]

I also remember SOHCAHTOA but Numberzz method seems more fun

Anyway thanks everyone. Good luck to anyone else on their exam period

 

PS: I don't think ya need to calculate the actual angle becuase its the non-calculator paper. Also it just says show that sin B = k/5 but thanks anyways

[Tim Smart]

5. So now I divide Opposite over Hypotenuse:

 

6. Right now we can consider that the x's cancel each other out becuase whatever number you put when you simplify the fractions you get:

 

Thats what I had so far. My question is that is the answer k = 4? And if so how would you explain the last step becuase you can't exactly say that they cancel each other out. Or can you?

 

Thanks

- Suhail

 

 

4x / 5x is the same as: 4 / 5 * x / x . Therefore you can cancel out the x's as any number over itself is equal to 1. So 4 / 5 * 1 = 4 / 5. That means you are correct in saying the x's cancel out.

 

 

I've graduated in Calculus so any algebra, differentiation, integration, trig etc. questions feel free to ask.

[Suhail]

Massive thanks to everyone who helped on the previous question. Trust me you guys and gals rock .

Right another question. This one is to do with Quadratics and Fractions + Equations on the Calculator Paper

 

The Question:

Solve the Equation:

 

 

Answer: x = ................................. or x = .................................

 

So we need to find the 2 values of x

 

1. Cross Multiply the fractions. That becomes:

2. Now simplify out the brackets:

3. Now multiply the left section to remove the fraction and simplify:

4. Now we want a quadratic so everything has to be on one side to make ax^2 + bx + c = 0:

5. Now to put it in the quadratic formula:

6. The answers of x comes out to be:

 

Are they the right answers for x because when I put them back in to the original equation it doesn't add up correctly? Have I done something Stupid?

 

Thanks

- Suhail

[mac_cute]

There is something wrong:take a nice look...

 

 

So we need to find the 2 values of x

 

Good so far

 

2. Now simplify out the brackets:

 

(x^2+9x-12)/(2x^2-x-3)=1 ->Here is the detail Suhail (bold). :angry2:

 

 

3. Now multiply the left section to remove the fraction and simplify:

 

x^2+9x-12=2x^2-x-3

 

4. Now we want a quadratic so everything has to be on one side to make ax^2 + bx +

 

x^2-10x+9=0

 

 

5. Now to put it in the quadratic formula:

 

 

6. The answers of x comes out to be:

 

x=9 or x=1

[Suhail]

Wow thanks XxP/TuX, I did the same working out as you on my piece of paper but one mistake I made was when I moved everything from right to left. 12--3 = 9 not 15 . Which is what I did.

 

Also I think that + x is a typo. Its meant to be - x. Whoops (How embarrasing :angry2: ). I really should learn how to use LaTeX and not Photoshop to write my equations.

 

Anyway massive thanks mate. . (Goes to finish off Completing the Square )

This is now solved

[mac_cute]

@Suhail No problem

 

@Tim Smart/All

 

Curiosity:there is any chance of integrate erf(x)?

[Tim Smart]

@Suhail No problem

 

@Tim Smart/All

 

Curiosity:there is any chance of integrate erf(x)?

 

I think so yes. Its been a while since integration.

 

 

Which is very intergratable. So yes is the answer

[mac_cute]

I think so yes. Its been a while since integration.

 

 

Which is very intergratable. So yes is the answer

 

Oh,I'm sorry because I was wrong. I mean erf(z) without any limit,an undefined integral like:

 

Integrate(z^2,z)=z^3/3.

 

So it will be

 

Integrate(e^(-z^2),z)=?.

 

Is this possible?