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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!

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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 $b$

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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

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

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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.

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

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