If BC is twice as long as AB, which equations would let you find the length of AC?
+118608 Is some of this question missing?
I am going to use the cosine rule.
$$\begin{array}{rll}
\overline{AC}^2&=&(2x)^2+x^2-(2*2x*x*Cos( \overline{AC}^2&=&4x^2+x^2-(4x^2Cos( \overline{AC}^2&=&5x^2-(4x^2Cos( \overline{AC}^2&=&x^2(5-4Cos( \overline{AC}&=&x\sqrt{5-4Cos( \overline{AC}&=&\overline{AB}\sqrt{5-4Cos( \end{array}$$
That should be okay if i didn't make any stupid mistakes.
+118608 Is some of this question missing?
I am going to use the cosine rule.
$$\begin{array}{rll}
\overline{AC}^2&=&(2x)^2+x^2-(2*2x*x*Cos( \overline{AC}^2&=&4x^2+x^2-(4x^2Cos( \overline{AC}^2&=&5x^2-(4x^2Cos( \overline{AC}^2&=&x^2(5-4Cos( \overline{AC}&=&x\sqrt{5-4Cos( \overline{AC}&=&\overline{AB}\sqrt{5-4Cos( \end{array}$$
That should be okay if i didn't make any stupid mistakes.
