# Find the other side lengths of a triangle given the perimeter, one side and an angle

I’m working through some maths questions and this one has me stumped. It is as follows: “The perimeter of triangle ABC = 15cm, given AB = 7cm and angle BAC = 60 degrees, find the lengths of AC and BC and the area of the triangle. If this was a right angled triangle then it’d be easy, but I’m not sure how to get the other sides if only one side, one angle and the perimeter are defined. Thanks!

Use that $$\sin(\frac{\alpha}{2})=\sqrt{\frac{(p-b)(p-c)}{bc}}$$ so we get $$\frac{c}{p-c}\times\sin^2(\frac{\alpha}{2})+1=\frac{p}{b}$$ and from here we get $$b$$

We can convert it to a problem of Loci... which may otherwise appear round about.

Since sum of remaining sides is given, the locus is a Newton's (planetary) ellipse... by virtue of ellipse property.

Since subtended angle $$\alpha$$ at opposite side is given, the locus is a circle.

The ellipse and circle cut at two points.

Let axes of ellipse be denoted in upper case, keeping lower case for the triangle..

We are given

$$a+b = 2 A = (a+b+c) - c , \quad 2 C = c$$

Eccentricity $$e = C/A$$

Minor axis $$B= A \sqrt{1-e^2}$$

Semi Latus-rectum $$p = B^2/A$$

Newton's ellipse in polar form

$$b = \frac {p}{1-e \cos \alpha}$$

Similarly find $$a$$ and then area using

$$\Delta= \sqrt {s(s-a)(s-b)(s-c)}..$$

By the Law of cosine you have $$BC^2=AB^2+AC^2-AB\cdot AC$$

Therefore $$AC+BC=8 \\ BC^2=49+AC^2-7 AC$$

This gives $$(8-AC)^2=49+AC^2-7 AC$$ which, after cancelation, is a linear equation.

• How have you got rid of Cos(A) at the end of the cosine rule here? – Nadim Oct 2 '19 at 18:13
• @Nadim $\cos(BAC)=\cos(60)=\frac{1}{2}$. The problem gives the angle is $60$. – N. S. Oct 2 '19 at 22:14

Law of cosines:

Let $$AB = 7$$, $$BC = x$$ and $$AC = 15-(7+x) = 8-x$$.

Let $$m\angle A = 60$$ and $$m\angle B = \theta$$ and $$m\angle C = 180-(m\angle B+60)=120 -\theta$$.

Law of cosines says:

$$AB^2 = BC^2 + AC^2 - 2BC*AC\cos m \angle C$$ or $$49=x^2+(8-x)^2 + 2x(8-x)\cos(120 -\theta)$$.

$$BC^2 = AB^2 + AC^2 -2AB*AC\cos m\angle A$$ or $$x^2=49+(8-x)^2 -2*7(8-x)\cos 60$$

and ....

Oh, forget the third one. This is the one we want! $$\cos 60 = \frac 12$$ so

$$x^2 = 49-(8-x)^2 - 2*7(8-x)*\frac 12$$

$$x^2 = -(x^2 -16x+64) + (7x -56)+49$$

$$x^2 = -x^2 +23x -71$$

$$2x^2 -23x + 71=0$$

$$x = \frac {23\pm \sqrt {23^2 - 4*2*71}}{4}=\frac {23 \pm \sqrt{473}}4$$

$$\sqrt{473}\approx 21.7$$ and $$\frac {23+21.7}4 > 8$$ so that's not possible.

So $$BC = x =\frac {23 \pm \sqrt{473}}4$$ and $$AC =8-\frac {23 \pm \sqrt{473}}4$$

And the area.

If we let $$AC=8-\frac {23 \pm \sqrt{473}}4$$ be the base. Then $$AB\sin m\angle A= 7\frac {\sqrt 3}2$$ is the height.

(or alternatively if $$AB$$ were the base then $$AC\sin m\angle A$$ is the height. Either way...)

Area is $$\frac 12 (8-\frac {23 \pm \sqrt{473}}4)*7\frac {\sqrt 3}2=$$

$$14 - 7\cdot\frac {23\sqrt 3\pm \sqrt {3*473}}{16}$$

.....

Now it's just a matter of figuring out where I made my obvious and inevitable arithmetic error.