Photo AI

At an activity centre, a zip-line, runs between two vertical poles, AB and CD, on level ground, as shown - Leaving Cert Mathematics - Question Question 1 - 2014

Question icon

Question Question 1

At-an-activity-centre,-a-zip-line,-runs-between-two-vertical-poles,-AB-and-CD,-on-level-ground,-as-shown-Leaving Cert Mathematics-Question Question 1-2014.png

At an activity centre, a zip-line, runs between two vertical poles, AB and CD, on level ground, as shown. The point E is on the ground, directly below the zip-line. ... show full transcript

Worked Solution & Example Answer:At an activity centre, a zip-line, runs between two vertical poles, AB and CD, on level ground, as shown - Leaving Cert Mathematics - Question Question 1 - 2014

Step 1

Find the distance |ED|, correct to one decimal place.

96%

114 rated

Answer

To find the distance |ED|, we can use the Pythagorean theorem.

Given:

  • |AE| = 12 m
  • |BE| = 14 m
  • |CD| = 1.95 m

We calculate |ED| as follows:

∣ED∣=sqrt∣AE∣2+∣CD∣2=sqrt(12)2+(1.95)2∣ED∣=sqrt144+3.8025∣ED∣=sqrt147.8025∣ED∣=12.1 m (to one decimal place)|ED| = \\sqrt{|AE|^2 + |CD|^2} = \\sqrt{(12)^2 + (1.95)^2} \\ |ED| = \\sqrt{144 + 3.8025} \\ |ED| = \\sqrt{147.8025} \\ |ED| = 12.1 \text{ m (to one decimal place)}

Step 2

Find |∠AEB|, correct to the nearest degree.

99%

104 rated

Answer

To find |∠AEB|, we can use the cosine rule.

Using:

  • |AE| = 12 m
  • |BE| = 14 m

We find |∠AEB| as follows:

cos⁡∣∠AEB∣=∣AE∣∣BE∣∣∠AEB∣=cos⁡−1(1214)∣∠AEB∣≈31° (to the nearest degree)\cos |∠AEB| = \frac{|AE|}{|BE|} \\ |∠AEB| = \cos^{-1}\left(\frac{12}{14}\right) \\ |∠AEB| ≈ 31° \text{ (to the nearest degree)}

Step 3

Find |∠DEB|, given that |∠CED| = 11°, correct to the nearest degree.

96%

101 rated

Answer

From the triangle, we know:

∣∠DEB∣=180°−(∣∠CED∣+∣∠AEB∣)∣∠DEB∣=180°−(11°+31°)∣∠DEB∣=180°−42°=138° (to the nearest degree)|∠DEB| = 180° - (|∠CED| + |∠AEB|) \\ |∠DEB| = 180° - (11° + 31°) \\ |∠DEB| = 180° - 42° = 138° \text{ (to the nearest degree)}

Step 4

Hence, or otherwise find the distance |DB|.

98%

120 rated

Answer

To find the distance |DB|, we can use the cosine rule again:

Using:

  • |DE| = 14 m
  • |EC| = 10 m
  • |∠DEB| = 138°

We find |DB| as follows:

∣DB∣=∣DE∣2+∣EC∣2−2⋅∣DE∣⋅∣EC∣⋅cos⁡∣∠DEB∣∣DB∣=(14)2+(10)2−2(14)(10)cos⁡(138°)∣DB∣=196+100−2⋅14⋅10⋅(−0.669)∣DB∣=296+187.38∣DB∣=483.38∣DB∣≈22.0 m (to one decimal place)|DB| = \sqrt{|DE|^2 + |EC|^2 - 2\cdot|DE|\cdot|EC|\cdot\cos |∠DEB|} \\ |DB| = \sqrt{(14)^2 + (10)^2 - 2(14)(10)\cos(138°)} \\ |DB| = \sqrt{196 + 100 - 2 \cdot 14 \cdot 10 \cdot (-0.669)} \\ |DB| = \sqrt{296 + 187.38} \\ |DB| = \sqrt{483.38} \\ |DB| ≈ 22.0 \text{ m (to one decimal place)}

Join the Leaving Cert students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;