Divide by long division method
(x
6 + 2x4+ 6x – 9 ) by (x
3 + 3)
Answers
Answer:
DBC________
Fig. 3.30
... angle BAC angleBDC (c.a.c.t.)
In A ABC, ZBAC = 60° . ZBDC = 60°
ZDAC = ZADC = ZACD = 60°... sum of angles of A ADC is 180°Kk
.: AADC is an equilateral triangle.
.. AC = AD = DC ........ corollary of converse of isosceles triangle theorem
1
But AB = AD........ construction
2
1
..AB AC ........ * AD= AC
2
1 / 2Proof : /\ ABC and /\ DBC
seg AB seg DB
___________
angle ABC segment Angle DBC________
seg BC seg BC_______
.. A ABC = A DBC________
Fig. 3.30
... angle BAC angleBDC (c.a.c.t.)
In A ABC, ZBAC = 60° . ZBDC = 60°
ZDAC = ZADC = ZACD = 60°... sum of angles of A ADC is 180°
.: AADC is an equilateral triangle.
.. AC = AD = DC ........ corollary of converse of isosceles triangle theorem
1
But AB = AD........ construction
2
1
..AB AC ........ * AD= AC
2
1 / 2Theorem: If the acute angles of a right angled triangle have measures 30° and 60°, then
the length of the side opposite to 30° angle is half the length of the hypotenuse,
(Fill in the blanks and complete the proof.)
Given : In A ABC
ZB = 90°, ZC = 30°, ZA= 60°
60
To prove : AB
-
AC
30
B
C С
Fig. 3.29
Construction : Take a point D on the extended
seg AB such that AB = BD. Draw seg DC.
Proof : /\ ABC and /\ DBC
seg AB seg DB
___________
angle ABC segment Angle DBC________
seg BC seg BC_______
.. A ABC = A DBC________
Fig. 3.30
... angle BAC angleBDC (c.a.c.t.)
In A ABC, ZBAC = 60° . ZBDC = 60°
ZDAC =
angle ABC segment Angle DBC________
seg BC seg BC_______
.. A ABC = A DBC________
Fig. 3.30DBC________
Fig. 3.30
... angle BAC angleBDC (c.a.c.t.)
In A ABC, ZBAC = 60° . ZBDC = 60°
ZDAC = ZADC = ZACD = 60°... sum of angles of A ADC is 180°Kk
.: AADC is an equilateral triangle.
.. AC = AD = DC ........ corollary of converse of isosceles triangle theorem
1
But AB = AD........ construction
2
1
..AB AC ........ * AD= AC
2
1 / 2Proof : /\ ABC and /\ DBC
seg AB seg DB
___________
angle ABC segment Angle DBC________
seg BC seg BC_______
.. A ABC = A DBC________
Fig. 3.30
... angle BAC angleBDC (c.a.c.t.)
In A ABC, ZBAC = 60° . ZBDC = 60°
ZDAC = ZADC = ZACD = 60°... sum of angles of A ADC is 180°
.: AADC is an equilateral triangle.
.. AC = AD = DC ........ corollary of converse of isosceles triangle theorem
1
But AB = AD........ construction
2
1
..AB AC ........ * AD= AC
2
1 / 2Theorem: If the acute angles of a right angled triangle have measures 30° and 60°, then
the length of the side opposite to 30° angle is half the length of the hypotenuse,
(Fill in the blanks and complete the proof.)
Given : In A ABC
ZB = 90°, ZC = 30°, ZA= 60°
60
To prove : AB
-
AC
30
B
C С
Fig. 3.29
Construction : Take a point D on the extended
seg AB such that AB = BD. Draw seg DC.
Proof : /\ ABC and /\ DBC
seg AB seg DB
___________
angle ABC segment Angle DBC________
seg BC seg BC_______
.. A ABC = A DBC________
Fig. 3.30
... angle BAC angleBDC (c.a.c.t.)
In A ABC, ZBAC = 60° . ZBDC = 60°
ZDAC =
angle ABC segment Angle DBC________
seg BC seg BC_______
.. A ABC = A DBC________
Fig. 3.30DBC________
Fig. 3.30
... angle BAC angleBDC (c.a.c.t.)
In A ABC, ZBAC = 60° . ZBDC = 60°
ZDAC = ZADC = ZACD = 60°... sum of angles of A ADC is 180°Kk
.: AADC is an equilateral triangle.
.. AC = AD = DC ........ corollary of converse of isosceles triangle theorem
1
But AB = AD........ construction
2
1
..AB AC ........ * AD= AC
2
1 / 2Proof : /\ ABC and /\ DBC
seg AB seg DB
___________
angle ABC segment Angle DBC________
seg BC seg BC_______
.. A ABC = A DBC________
Fig. 3.30
... angle BAC angleBDC (c.a.c.t.)
In A ABC, ZBAC = 60° . ZBDC = 60°
ZDAC = ZADC = ZACD = 60°... sum of angles of A ADC is 180°
.: AADC is an equilateral triangle.
.. AC = AD = DC ........ corollary of converse of isosceles triangle theorem
1
But AB = AD........ construction
2
1
..AB AC ........ * AD= AC
2
1 / 2Theorem: If the acute angles of a right angled triangle have measures 30° and 60°, then
the length of the side opposite to 30° angle is half the length of the hypotenuse,
(Fill in the blanks and complete the proof.)
Given : In A ABC
ZB = 90°, ZC = 30°, ZA= 60°
60
To prove : AB
-
AC
30
B
C С
Fig. 3.29
Construction : Take a point D on the extended
seg AB such that AB = BD. Draw seg DC.
Proof : /\ ABC and /\ DBC
seg AB seg DB
___________
angle ABC segment Angle DBC________
seg BC seg BC_______
.. A ABC = A DBC________
Fig. 3.30
... angle BAC angleBDC (c.a.c.t.)
In A ABC, ZBAC = 60° . ZBDC = 60°
ZDAC =
angle ABC segment Angle DBC________
seg BC seg BC_______
.. A ABC = A DBC________
Fig. 3.30