How to solve this similar triangle question?

Question

Isosceles triangle ABC satisfies: $|AB|=|AC|=5$, and $\angle BAC>60$ degrees. The length of the perimeter of this triangle is expressed with a whole number. How many triangles of that kind are there?

   A. 1
   B. 2
   C. 3
   D. 4
   E. 5

Answer 1

HINT:

$$\cos\angle BAC=\frac{AB^2+AC^2-BC^2}{2 AB\cdot AC}$$

and as $\displaystyle\angle BAC>60^\circ,\cos\angle BAC<\frac12$

Establish that $|BC|>5$

Now use Sum of Two Sides of Triangle Greater than Third Side

Answer 2

Hint:

By simple trigonometry, the third side $BC$ has length $2\times 5 \times \sin(\frac{1}{2}\angle BAC)$, which is $10 \times \sin(\frac{1}{2}\angle BAC)$.

As $\angle BAC$ is greater than $60^\circ$, what value will length $|BC|$ exceed?

Now, what do you think length $|BC|$ should be less than? Hint: the Triangle Inequality.

Once you have the bounds for length $|BC|$, how many integer values lie in these bounds?

Here is a picture to help explain things:-

You can use the Sine Rule to calculate $|BC|$ - although it's a bit more involved, I will go through it. We have

$\large \frac{|BC|}{\sin \angle BAC}=\frac{5}{\sin \angle CBA}$

Now, as ABC is an isosceles triangle, $\angle CBA= 0.5\times(180^{\circ}-\angle BAC)= 90^{\circ}-\frac{1}{2}\angle BAC$

So, we have

$\large \frac{|BC|}{\sin \angle BAC}=\frac{5}{\sin (90^{\circ}-\frac{1}{2}\angle BAC)}=\frac{5}{\cos \frac{1}{2}\angle BAC}$

Using the double angle formula, $\sin \angle BAC=2sin(\frac{1}{2}\angle BAC)cos(\frac{1}{2}\angle BAC)$, resulting in

$\large \frac{|BC|}{2sin(\frac{1}{2}\angle BAC)cos(\frac{1}{2}\angle BAC)}=\frac{5}{\cos \frac{1}{2}\angle BAC}$

leading to

$|BC|=2\times 5 \times \sin(\frac{1}{2}\angle BAC)$