Math trouble. can you help?

Question

i dont know how to do this problem. its very very hard. = )

#2. a farmer owns a triangular field ABC. The side [AC] is 104m, the side [AB] is 65m and the angle between these two sides is 60 degrees.

a) calculate the length of the third side of the field

b) find the area of the field in the form p *squareroot*(3), where p is an integer

(same prob) Let D be a point on [BC] such that [AD] bisects the 60 degree angle. The farmer divides the field into two parts by constructing a straight fence [AD] of length x meters.

c) (I) show that the area of the smaller part is given by (65x) / 4 and find an expression for the area of the larger part
(II) hence, find the value of x in the form q *squareroot* (3), where q is an integer

d) prove that BD / DC = 5 / 8

PS. if its too long to answer here, you can email me at [e-mail]

thank again.

Answer 1

Have you learned trig? It is very easy once you are familiar with trig.
a) Using Cosine Law, we have:
[BC]^2 = [AB]^2 + [AC]^2 -2[AB][AC]Cos A
= 65^2 + 104^2 - 65*104
= 4225 + 39*104 = 8281
Hence [BC] = 91 (m)
b) Remember the area of a triangle is 0.5 ab Sin C:
Area = (104/2)65 Sin A
= 52*65*Sqrt(3)/2
= 1690 Sqrt(3) (m^2)
d) Let us solve (d) first.
Since AD is the angular bisector, so
BD/DC = AB/AC = 65/ 104 = 5/8
c) Consider BC is the base of the triangle, so the area of ADB ratio the area of ABC = BD/BC = 5/13. Therefore,
the area of ADB = (5/13) (Area of ABC)
= 650 sqrt(3) (m^2),
and the value of x is 40sqrt(3)
The area of the larger part is
1690 sqrt(3) - 650 sqrt(3) = 1040 sqrt(3) (m^2)