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
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
#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.
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)
$