In die diagram is B, E en D punte in dieselfde horizontale vlak. AB en CD is vertikale pale. Staalkabels AE en CE anker die pale by E. 'n Ander staal kabel verbind A en C. CE = 8,6 m, BE = 10 m, \angle AEB = 40^{\circ}, \angle AEC = 70^{\circ} en \angle CED = 27^{\circ}. Bereken die: 1. Hoogte van paal CD 2. Lengte van kabel AE 3. Lengte van kabel AC

Photo AI

In die diagram is B, E en D punte in dieselfde horizontale vlak - NSC Mathematics - Question 7 - 2017 - Paper 2

Question 7

In die diagram is B, E en D punte in dieselfde horizontale vlak. AB en CD is vertikale pale. Staalkabels AE en CE anker die pale by E. 'n Ander staal kabel verbind A... show full transcript

Worked Solution & Example Answer:In die diagram is B, E en D punte in dieselfde horizontale vlak - NSC Mathematics - Question 7 - 2017 - Paper 2

Step 1

Hoogte van paal CD

96%

114 rated

Answer

To find the height of pole CD, we apply the sine function in the following equation:

$\sin(27^{\circ}) = \frac{CD}{8.6}$

Rearranging gives: $CD = 8.6 \cdot \sin(27^{\circ})$

Calculating this results in: $CD \approx 3.90 m$ .

Step 2

Lengte van kabel AE

99%

104 rated

Answer

We can use the cosine function for this cable length:

$\cos(40^{\circ}) = \frac{10}{AE}$

Rearranging gives: $AE = \frac{10}{\cos(40^{\circ})$

Calculating this gives: $AE \approx 13.05 m$ .

Step 3

Lengte van kabel AC

96%

101 rated

Answer

Using the cosine rule in triangle AEC, we can calculate AC:

$AC^2 = CE^2 + AE^2 - 2 \cdot CE \cdot AE \cdot \cos(70^{\circ})$

Substituting the known values yields: $AC^2 = (8.6)^2 + (13.05)^2 - 2 \cdot (8.6)(13.05)(\cos(70^{\circ}))$

Calculating gives: $AC^2 \approx 167.49$ Thus, AC is: $AC \approx 12.94 m$ .

Join the NSC students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;