If the sides of a triangle satisfy $(a-c)(a+c)^2+bc(a+c)=ab^2$, and if one angle is $48^\circ$, then find the other angles.

Question

In triangle $ABC$ one angle of which is $48^{\circ}$, length of the sides satisfy the equality: $$(a-c)(a+c)^2+bc(a+c)=ab^2$$ Find the value in degrees the other two angles of the triangle.

I have no idea how to solve this problem

Answer 1

$$(a-c)(a+c)^2+bc(a+c)=ab^2$$ $$\implies a^3-a b^2-a c^2+b c^2+a^2 c+a b c-c^3=0$$ $$\implies a(a^2-b^2)-c^2(a-b)+ca(a+b)-c^3=0$$ $$\implies a(a+b)(a-b)-c^2(a-b)+ca(a+b)-c^3=0$$ $$\implies (a-b)(a(a+b)-c^2)+c(a(a+b)-c^2)=0$$ $$\implies (a-b+c)(a(a+b)-c^2)=0$$ Dividing the both sides by $a-b+c\gt 0$ gives $$c^2=a^2+ab$$ from which we have $c\gt a$.

Case 1 : $c^2=a^2+b^2$

$a=b\implies c=\sqrt 2\ a$, which contradicts that the triangle has $48^\circ$.

Case 2 : $c^2\gt a^2+b^2$.

$a\gt b\implies C\gt 90^\circ, A=48^\circ, B\lt 42^\circ$.

By the law of cosines, $$a^2=b^2+a^2+ab-2b\sqrt{a^2+ab}\ \cos(48^\circ)\implies \frac ba=4\cos^2(48^\circ)-1$$ Also, $$c^2=a^2+ab=a^2+b^2-2ab\cos C\implies \cos C=\frac{b}{2a}-\frac 12=2\cos^2(48^\circ)-1=\cos(96^\circ)$$ Therefore, $A=48^\circ,B=36^\circ,C=96^\circ$.

Case 3 : $c^2\lt a^2+b^2$.

$a\lt b$ and $C\lt 90^\circ$.

If $B=48^\circ$, then $$\begin{align}&b^2=a^2+a^2+ab-2a\sqrt{a^2+ab}\ \cos(48^\circ)\\&\implies 2\cos(48^\circ)=\sqrt{2\cos C+2}\ (1-2\cos C)\\&\implies 2\cos(48^\circ)=2\cos\frac{C}{2}\left(1-2\left(2\cos^2\frac{C}{2}-1\right)\right)\\&\implies\cos(48^\circ)=3\cos\frac C2-4\cos^3\frac C2\\&\implies\cos(48^\circ)=-\cos\left(\frac{3C}{2}\right)\\&\implies C=32^\circ,88^\circ\end{align}$$

If $C=48^\circ$, then $$c^2=a^2+ab=a^2+b^2-2ab\cos(48^\circ)\implies \cos(48^\circ)=\frac 12(4\cos^2A-1)-\frac 12=\cos(2A)$$$$\implies A=24^\circ$$

Therefore, the answer is $$\color{red}{(A,B,C)=(48^\circ, 36^\circ, 96^\circ),(44^\circ,48^\circ, 88^\circ),(24^\circ,108^\circ,48^\circ)}$$

Answer 2

From $c^2-a^2=ab,$

using $(i)$Sine Law,

$(ii)$Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

and $(iii)\sin(C+A)=\cdots=\sin B$

$$\sin B\sin(C-A)=\sin A\sin B$$

As $\sin B\ne0,\implies\sin(C-A)=\sin A$

$\implies C-A=n\pi+(-1)^nA$ where $n$ is any integer

If $n$ is odd, $n=2m+1$(say) $\implies C-A=(2m+1)\pi+(-1)A$

$\iff C=(2m+1)\pi$ which is impossible as $0<C<\pi$

If $n$ is even, $n=2m$(say) $\implies C-A=2m\pi+A\iff C=2m\pi+2A$

As $0<C,A<\pi; C=2A$

We also have $A+B+C=\pi$

Now check for the three cases namely, $A=48^\circ, B=48^\circ,C=48^\circ,$