A triangle with an angle of $60^\circ$ and side lengths $5$, $2x-3$, and $x+1$ ...

Question

I've been given the following:.

I don't understand what part a is asking me to do. My initial assumption is to factorise the equation to find $x$, even though part b already asks for that.

Thanks for reading.

Answer 1

How would you solve for sides given only an angle?

The cosine rule!

In this case if we were to concentrate it on $AC$ then:

$(x+1)^2=5^2+(2x-3)^2-2 \cdot 5(2x-3)\cos 60^\circ$

$x^2+2x+1=25+4x^2-12x+9-(10x-15)$

$3x^2-24x+48=0 \implies x^2-8x+16=0$

I think that's it.

For c,

Use the sine rule

Answer 2

As saulspatz mentioned in his comment you can use the law of cosines.

In the given triangle $ABC$ we have

\begin{align} \overline{AC}^2 = \overline{AB}^2 + \overline{BC}^2 - 2\cdot \overline{AB} \cdot \overline{BC} \cdot \cos(\angle ABC) \end{align}

and with the given numbers that means

\begin{align} (x+1)^2 = 5^2 + (2x-3)^2 - 2 \cdot 5 \cdot (2x-3) \cdot \cos(60°). \end{align}

Since $\cos(60°) = \dfrac{1}{2}$ we have

\begin{align} (x+1)^2 = 5^2 + (2x-3)^2 - 5 \cdot (2x-3). \end{align}

We can now use the first and second binomial formulas and receive

\begin{align} x^2 + 2x + 1 = 25 + 4x^2 - 12x + 9 - 10 x + 15 \end{align}

which can be simplified

\begin{align} 0 = 3x^2 - 24x + 48. \end{align}

Finally, dividing by $3$ yields

\begin{align} 0 = x^2 - 8x + 16. \end{align}