1. ## trig triangles

A farmer owns a triangular field ABC. One side of the triangle, AC, is 104 m, a second side, AB, is 65 m and the angle between the two sides is 60 degrees.

(a) Given that sin60 = √3/2, find the area of the field in the form p√3 wehre p is an integer.

Let D be a point on BC such that AD bisects the 60 degree angle. The farmer divides the field into two parts A1 and A2 by constructing a straight fence AD of length x meters.

(b) (i) show that the area of A1 is given by 65x/4.
(ii) Find a similar expression for the area of A2.
(iii) hence, find the value of x in the form q√3, where q is an integer.

(c) (i) explain why sinA(angle D)C = sinA(angle D)B.
(ii) use the result of part (i) and the sine rule to show that BD/DC = 5/8

thanks

2. (a)
$\displaystyle \displaystyle A=\frac{AB\cdot AC\cdot\sin A}{2}=\frac{65\cdot 104\cdot\sin 60}{2}=\frac{65\cdot 104\cdot\frac{\sqrt{3}}{2}}{2}=1690\sqrt{3}$

(b)
i. $\displaystyle \displaystyle A_1=\frac{AB\cdot AD\cdot\sin\widehat{BAD}}{2}=\frac{65\cdot x\cdot\frac{1}{2}}{2}=\frac{65x}{4}$
ii. $\displaystyle \displaystyle A_2=\frac{AC\cdot AD\cdot\sin\widehat{CAD}}{2}=\frac{104\cdot x\cdot\frac{1}{2}}{2}=26x$
iii. $\displaystyle A=A_1+A_2\Rightarrow 1690\sqrt{3}=\frac{65x}{4}+26x\Rightarrow x=40\sqrt{3}$

(c)
i. $\displaystyle \widehat{ADB}+\widehat{ADC}=180\Rightarrow\sin\wid ehat{ADB}=\sin (180-\widehat{ADC})=\sin\widehat{ADC}$
ii. $\displaystyle \displaystyle \frac{BD}{\sin 30}=\frac{AB}{\sin\widehat{ADB}}\Rightarrow BD=\frac{65}{2\sin\widehat{ADB}}$
$\displaystyle \displaystyle \frac{DC}{\sin 30}=\frac{AC}{\sin\widehat{ADBC}}\Rightarrow CD=\frac{104}{2\sin\widehat{ADB}}$
So $\displaystyle \displaystyle \frac{BD}{DC}=\frac{65}{104}=\frac{5}{8}$