1. ## Solid trigonometry issue..

I have attached the question and the diagram. I was able to find answer for a, but need help with the rest.

I just can't seem to figure out how to solve b. As you see, this is a GCSE level question, you'll have to be a bit basic here.

Thanks

2. #b Is angle EFH, which is 90 - angle CFH.

Consider the right triangle CFH, CH is 10 and FH is 20, both given:

sin CFH = 10/20 = 1/2

So, 90 - sin^-1 (1/2) =

For # c and # d.

Angle ACB = BAC = 70 degrees

The length line from the mid-point of AC to the mid-point of GH is the length of the line from the mid-point of AC to B minus the length of the line from the mid-point GH to B:

14sin(70) - 4sin(70) = 9.397

Together with the line joining the mid-points of AC and DF, 10sqrt(3) = 17.321, from #a, and the line joining the mid-points of DF and GH you have a right triangle.

#c tan^-1 (17.321/9.397) =

#d Using the same triangle: (9.387)^2 + (17.321)^2) = 3888.302

sqrt 388.302 =

I hope this helps. I hope it's correct.

3. Thanks. But I still didn't understand #c and #d ... I just don't get what I have to do there despite wasting another hour on it ...

4. Sorry that didn't help. Does this.

Use the white right triangle. Angle C is for #c, it corresponds to the one asked for.

Then find the hypotenuese for #d.

5. Thanks a lot got it now.

This part is new to me though:

14sin(70) - 4sin(70) = 9.397

On what basis can we do this. Is it because GB is 4cm in length and GH || AC?

6. Triangle ABC is isosceles so its height, the perpendicular from the midpoint of AC to the vertex B, is given by:

$\displaystyle sin(70) = \frac{Opp}{Hyp}$

$\displaystyle sin(70) = \frac{Opp}{14}$

14 * sin(70) = Opp

Yes, triangles ABC and BGH are similar so the line joining the midpoints of AC and GH is:

$\displaystyle 14 * sin(70) - \frac{4}{14} * 14 * sin(70) = 14 * sin(70) - 4 * sin(70)$