1 58 F-rule
2 24 Z-rule
3 98 Angle sum of a triangle is 180°
4 58 Equal angles as a result of an isosceles triangle
5 64 Angle sum of a triangle is 180°
6 64 Angle sum of a triangle is 180°
7 64 Z-rule
8 71 Angle sum of a triangle is 180°
9 109 C-rule
b)
1 45° C-rule
2 100° X-rule
3 35° Angle sum of a triangle is 180°
4 35° C-rule
5 35° Z-rule
6 35° Z-rule
7 80° X-rule
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20 / 10 = 2
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2) Given two similar triangles, all corresponding sides are equally proportional.
Solve for M and N:
Solve for M and N:
M = 14
N = 8
20 / 10 = 2
7 x 2 = 14
16 / 2 = 8
3a) Solve the missing angles:
1 26°
2 72°
3 53°
4 53°
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1 94°
5 214°
b)
1 94°2 94°
3 86°
4 45°
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2) x² = x² - 20x + 100 + 25
3) 20x = 325
A) What is the length of the arc of the watered sector, to the nearest metre?
The Length, l, of and arc with a central angle of m° is given my the formula:
1) l = m
--- x 2πr
360
2) l = 100
----- x 2π x 25
360
3) l = 44 m°
B) What is the area of the watered sector, to the nearest square metre?
The area, A, of a sector with a central angle of m° is given by the formula:
4) A hemispherical pot is used for a hanging basket. The width of the surface of the soil is 30cm. The maximum depth of the soil is 10cm. Fine the radius of the pot:
1) x² = (x-10)² + 15²
2) x² = x² - 20x + 100 + 25
3) 20x = 325
4) ---- -----
20 20
5) x = 16.25
5) x = 16.25
The radius of the pot is 16.25cm.
5) A water sprinkler turns back and forth through an angle of 100° . The water sprays out to a maximum distance of 25m.
A) What is the length of the arc of the watered sector, to the nearest metre?
The Length, l, of and arc with a central angle of m° is given my the formula:
1) l = m
--- x 2πr
360
2) l = 100
----- x 2π x 25
360
3) l = 44 m°
B) What is the area of the watered sector, to the nearest square metre?
The area, A, of a sector with a central angle of m° is given by the formula:
1) A = m°
----- x π r²
360
2) A = 100
----- x π x 25²
360
3) A = 545 m²
