Typical problems of mathematical trigonometric functions

Solution: (1) Because the quadrilateral ABCD is an inscribed quadrilateral of a circle,

So angle A+ angle C= 180 degrees,

So COSA =-COSC,

According to cosine theorem, BD 2 = AB 2+DA 2-2 xadxdaxosa.

bd^2=bc^2+cd^2-2xbcxcdxcos，

So1+4-4 COSA = 4+9-12 COSC.

5+4 Xhosa = 13- 12 Xhosa

16cosC=8

cosC= 1/2

Angle C=60 degrees.

So BD 2 = 4+9- 12x (1/2).

BD= root number 7.

(2) Because the angle C=60 degrees,

So the angle A= 120 degrees,

So the area of the triangle ABD = (1/2) xabxasina.

=( 1/2)x 1x2x (root number 3)/2

= (root number 3)/2,

Area of triangle BCD =( 1/2)xBCxCDxsinC

=( 1/2)x2x3x (root number 3)/2

=(3 radicals 3)/2,

So the area of quadrilateral ABCD = the area of triangle ABD+the area of triangle BCD.

The root number 3 of =2.

According to sine theorem:

Circumferential circle diameter =BD/sinC

= (root number 7)/[ (root number 3)/2]

=(2 radicals 2 1)/3.