A flagpole [GH], shown in the diagram, is vertical and the ground is inclined at an angle of 5° to the horizontal between E and G. The angles of elevation from E and F to the top of the pole are 35° and 52° respectively. The distance from E to F along the incline is 6 m. Find how far F is from the base of the pole (G) along the incline. Give your answer correct to two decimal places. In the diagram the large circle s has centre O and the small circle c has centre D. The circle c touches the circle s at the point C. OA and OB are tangents to c as shown. The radius of s is r. ∠LBOA1 = 60°. The ratio of the area of s to the area of c is k : 1. Find the value of k.

Leaving Cert Mathematics Trigonometry: A flagpole [GH], shown in the di

A flagpole [GH], shown in the diagram, is vertical and the ground is inclined at an angle of 5° to the horizontal between E and G - Leaving Cert Mathematics - Question 3 - 2020

Question 3

A-flagpole-[GH],-shown-in-the-diagram,-is-vertical-and-the-ground-is-inclined-at-an-angle-of-5°-to-the-horizontal-between-E-and-G-Leaving Cert Mathematics-Question 3-2020.png

A flagpole [GH], shown in the diagram, is vertical and the ground is inclined at an angle of 5° to the horizontal between E and G. The angles of elevation from E and... show full transcript

Worked Solution & Example Answer:A flagpole [GH], shown in the diagram, is vertical and the ground is inclined at an angle of 5° to the horizontal between E and G - Leaving Cert Mathematics - Question 3 - 2020

Step 1

Find how far F is from the base of the pole (G) along the incline.

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Answer

To find the distance from F to G along the incline, we can use trigonometric ratios based on the angles of elevation.

Using Triangle EFG:
- Using the angle of elevation from E (35°), we have:
  
  $an(35°) = \frac{GH}{EG}$
- Rearranging gives:
  
  $GH = EG \cdot an(35°)$
Using Triangle FGH:
- Using the angle of elevation from F (52°), we have:
  
  $an(52°) = \frac{GH}{FG}$
- Rearranging gives:
  
  $GH = FG \cdot an(52°}$
Substituting the distance from E to F (EG = 6 m):
- We can use the sine rule to connect FG and the distances:
  
  $FG = 6 m \cdot \sin(5°)$
Calculating the height GH:
- For base G, substituting known values:
  
  $GH = 6 m \cdot an(35°)$
  
  Calculating this gives:
  
  $GH \approx 4.81 m$
Finding distance F to G:
- Now substitute to find the distance from F to G:
  
  $d = GH \div \sin(5°)$
  
  Solving this will give the final distance along the incline.

Step 2

Find the value of k.

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Answer

Finding the area of the large circle (s):
- The area of a circle is given by:
  
  $\text{Area} = \pi r^2$
- Thus, for circle s with radius r:
  
  $\text{Area}_s = \pi r^2$
Finding the radius of the small circle (c):
- From the angle ∠LBOA1 = 60°, we can deduce that
  
  $\sin(30°) = \frac{r_c}{r}$
- Thus,
  
  $r_c = \frac{r}{2}$
Finding the area of the small circle (c):
- Using the radius derived:
  
  $\text{Area}_c = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}$
Ratio of areas:
- Given the ratio of areas is k : 1:
  
  $k = \frac{\text{Area}_s}{\text{Area}_c} = \frac{\pi r^2}{\frac{\pi r^2}{4}} = 4$
Final answer:
- Therefore, the value of k is 4.

A flagpole [GH], shown in the diagram, is vertical and the ground is inclined at an angle of 5° to the horizontal between E and G - Leaving Cert Mathematics - Question 3 - 2020

Worked Solution & Example Answer:A flagpole [GH], shown in the diagram, is vertical and the ground is inclined at an angle of 5° to the horizontal between E and G - Leaving Cert Mathematics - Question 3 - 2020

Find how far F is from the base of the pole (G) along the incline.

Find the value of k.

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