# Thread: GCSE maths exam tomorrow and in desperate need of help...:)

1. ## GCSE maths exam tomorrow and in desperate need of help...:)

I was working on mymaths and just copied and pasted a few problems that I couldn't understand. The main one is 3D trigonometry and finding angles between planes.

1. Finding angles with the same 'sin' value but when the curve is reflected in the x axis.

2. 3D - finding an angle between a line and a plane.

3. Similar question to above.

4. Creating simultaneous equations from a graph.

5. Volume question.

6. Volume question.

Sorry there are so many!

Thanks

2. Hi Turple,

1) For point A remember that the curve is $y=-2\sin x$ this means that the curve $y=-\sin x$ has been stretched by scale factor 2 parrallel to the y-axis (i.e. all the y co-ordinates on the curve $y=-\sin x$ have been multiplied by 2)

For the equation we have;

$-2\sin x~=~0.4$

$\sin x~=~-0.2$

Well $\sin^{-1}(0.2)~\approx~11.54$

Thus due to the graph or CAST diagram or whatever method has been taught;

$x~=~180+11.54~,~360-11.54$

$x~=~191.54~,~348.46$

2) For 2 you have the length BE so imagine the triangle that BEH makes.

Notice that the angle at E is a right-angle thus trigonometry can be used.

We want the angle at B. We have length BE which is adjacent and we have length EH which is opposite hence use $\tan$

I get the answer to be $62.23$ to 2 decimal places.

3) As stated this question is similar so the information above should help.

4) In this question you need to use the 2 points that are given and sub them into the equation given; $y~=~ax^2+b$

We know that when $x~=~-3~,~y~=~-9$

Thus giving us;

$-9~=~(-3)^2a+b~~~[1]$

We also know that when $x~=~2~,~y~=~-14$

Hence

$-14~=~(2)^2a+b~~~[2]$

Now solve $[1]$ and $[2]$ simultaneously.

I get $a~=~1~~,~~b~=~-18$

5) We know the volume of a cone is $\frac{1}{3}\pi r^2 h$

So firstly we need to find the area of the larger cone then the smaller cone then minus one from the other to get the volume of the cup.

The question gives the area of the larger cone thus;

$272~=~\frac{1}{3}\pi r^2 h$

We know the height of the larger cone is $2h$ hence;

$272~=~\frac{1}{3}\pi r^2 (2h)$

To work out the volume of the smaller cone we need the radius so re-arrange the above formula. I get;

$r^2~=~\frac{408}{\pi h}$

Now we have the square of the radius of the smaller cone and the height of the smaller cone which is h. Sub the information into the cone formula;

$V~=~\frac{1}{3}\pi r^2 h$

$\implies~~V~=~\frac{1}{3}\pi (\frac{408}{\pi h})h$

The $h$ and the $\pi$ will cancel out giving the volume to be $136cm^3$

Now minus one from the other.

6) This one is similar to the one above. The equation for the volume of a cylinder is $V~=~\pi r^2 h$

3. Thankyou for your help but I don't understand how the two equations were solved simultaneously?

4. We have;

$-9~=~(-3)^2a+b~~[1]$

$-14~=~(2)^2a+b~~[2]$

Now this can be done several ways depending on preference. I would for these two make $b$ the subject of both the equations.

$[1]~~-9~=~(-3)^2a+b~~$

$\implies~-9~=~9a+b$

Minus the $9a$ from both sides,

$-9-9a~=~9a-9a+b$

$\implies~b~=~-9-9a$

$[2]~~-14~=~(2)^2a+b~~$

$\implies~-14~=~4a+b$

Minus $4a$ from both sides,

$-14-4a~=~4a-4a+b$

$\implies~b~=~-14-4a$

We can now say that,

$-14-4a~=~-9-9a$

Add $9a$ to both sides,

$-14-4a+9a~=~-9-9a+9a$

$\implies~-14+5a~=~-9$

Now add $14$ to both sides,

$-14+14+5a~=~-9+14$

$\implies~5a~=~5$

Divide both sides by $5$ to get $a~=~1$

Now sub in the value of a into either $[1]$ or $[2]$ to get the value for $b$.

Hope this helps.

5. Thankyou that has really helped

Fingers crossed for my exam today!