Question

Divide (

3 −216) by (y – 6 )and find the remainder by synthetic division​

Answer 1

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. 3.30