Question

IF xcosA=ycos(A+120)=zcos(A+240) then find xy+yz+xz

Answer 1

change the values to diff quadrant

Answer 2

The value of xy+yz+xz=0.  

Given:

$x \cos A=y \cos (A+120)=z \cos (A+240)$

To find the value of xy+yz+xz

Using the given expression  

Reduce the angles using allied angles formula so it will be easy to substitute the exact values of θ  

The formula of allied angles are  

$\begin{aligned} \cos (90+\theta) &=-\sin \theta \\ \cos (180+\theta) &=-\cos \theta \end{aligned}$

Using this two forms in the y and z terms  

$\begin{aligned} x \cos A &=y \cos (A+120)=z \cos (A+240) \\ x \cos A &=y \cos (90+30+A)=z \cos (180+60+A) \end{aligned}$

Comparing with the formula the expression becomes  

$\begin{array}{l}{x \cos A=-y \sin (A+30)=-z \cos (A+60)} \\ {x \cos A=-y \sin (A+30)=-z \cos (A+60)=a(\text { say })}\end{array}$

∴ x cos⁡A = a  

$\begin{array}{l}{\Rightarrow \frac{a}{x}=\cos A \quad \rightarrow(1)} \\ {-y \sin (A+30)=a} \\ {\Rightarrow \frac{a}{y}=-\sin (A+30) \rightarrow(2)} \\ {-z \cos (A+60)=a} \\ {\Rightarrow \frac{a}{z}=-\cos (A+60) \rightarrow(3)}\end{array}$

Adding equation (1), (2) and (3)

$\frac{a}{x}+\frac{a}{y}+\frac{a}{z}=\cos A-\sin (A+30)-\cos (A+60)$

Using trigonometry formula  

$\sin (a+b)=\sin a \cos b+\cos a \sin b$

And  

$\cos (a+b)=\cos a \cos b-\sin a \sin b$

Substituting the above two formula in the second and third term of right side  

We get,

$a \frac{(x y+y z+x z)}{x y z}=\cos A-\sin A \cos 30-\cos A \sin 30-\cos A \cos 30+\sin A \sin 60$

The value of $\sin 30 \text { is } \frac{1}{2}$

The value of $\cos 30 \text { and } \sin 60 \text { is } \frac{\sqrt{3}}{2}$

Substituting the values of sin30, cos⁡30 and sin⁡60

$a \frac{(x y+y z+x z)}{x y z}=\cos A-\sin A \frac{\sqrt{3}}{2}-\frac{\cos A}{2}+\frac{\cos A}{2}+\sin A \frac{\sqrt{3}}{2}$

$\begin{array}{c}{a \frac{(x y+y z+x z)}{x y z}=\cos A-\frac{\cos A}{2}-\frac{\cos A}{2}} \\ {a \frac{(x y+y z+x z)}{x y z}=\cos A-\cos A} \\ {a \frac{(x y+y z+x z)}{x y z}=0}\end{array}$

xy+yz+xz=0 .