MHB Finding $\angle ACB$ in $\triangle ABC$

[Albert1]

$\triangle ABC ,\angle ABC =45^o,\,\,point \,\, D \,\, on \,\, \overline{BC}, \,\,and :$
$2\overline {BD}=\overline {CD},\,\,\angle DAB=15^o$
please find :$\angle ACB=?$

 

[anemone]

My solution:

Spoiler

This is most likely not what you're looking for but I couldn't resist to solve any given geometry problem with the help of my close friend, aka trigonometry, hehehe...:p

View attachment 3568

First let $BD=DE=EC=k$ and join DG and GF such that $\angle ABC=45^{\circ}=\angle DGB$, $\angle GDF=30^{\circ}=\angle DFG$ and $\angle DAB=15^{\circ}=\angle AGF$ so triangles $DBG$, $DGF$ and $AGF$ are isosceles with the congruent sides labeled as $k$.

Applying the Sine Rule to the triangle $DFG$ we have

$\dfrac{DF}{\sin 120^{\circ}}=\dfrac{k}{\sin 30^{\circ}}\,\,\,\implies DF=\sqrt{3}k$

Applying the Cosine Rule to the triangle $ACD$ we get

$AC^2=(2k)^2+((1+\sqrt{3})k)^2-2(2k)(1+\sqrt{3})k)\cos 60^{\circ}\,\,\,\implies AC=\sqrt{6}k$

Applying the Sine Rule again to the triangle $ACD$ we see that

$\dfrac{AC}{\sin 60^{\circ}}=\dfrac{AD}{\sin ACB}$

$\sin ACB=\dfrac{1+\sqrt{3}}{2\sqrt{2}}=\dfrac{1}{2}\cdot\dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{\sqrt{2}}=\sin 30^{\circ}\cos 45^{\circ}+\cos 30^{\circ}\sin 45^{\circ}=\sin(30^{\circ}+45^{\circ})=\sin 75^{\circ}$

$\therefore \angle ACB=75^{\circ}$.

 

[Albert1]

My solution:

Spoiler

This is most likely not what you're looking for but I couldn't resist to solve any given geometry problem with the help of my close friend, aka trigonometry, hehehe...:p

View attachment 3568

First let $BD=DE=EC=k$ and join DG and GF such that $\angle ABC=45^{\circ}=\angle DGB$, $\angle GDF=30^{\circ}=\angle DFG$ and $\angle DAB=15^{\circ}=\angle AGF$ so triangles $DBG$, $DGF$ and $AGF$ are isosceles with the congruent sides labeled as $k$.

Applying the Sine Rule to the triangle $DFG$ we have

$\dfrac{DF}{\sin 120^{\circ}}=\dfrac{k}{\sin 30^{\circ}}\,\,\,\implies DF=\sqrt{3}k$

Applying the Cosine Rule to the triangle $ACD$ we get

$AC^2=(2k)^2+((1+\sqrt{3})k)^2-2(2k)(1+\sqrt{3})k)\cos 60^{\circ}\,\,\,\implies AC=\sqrt{6}k$

Applying the Sine Rule again to the triangle $ACD$ we see that

$\dfrac{AC}{\sin 60^{\circ}}=\dfrac{AD}{\sin ACB}$

$\sin ACB=\dfrac{1+\sqrt{3}}{2\sqrt{2}}=\dfrac{1}{2}\cdot\dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{\sqrt{2}}=\sin 30^{\circ}\cos 45^{\circ}+\cos 30^{\circ}\sin 45^{\circ}=\sin(30^{\circ}+45^{\circ})=\sin 75^{\circ}$

$\therefore \angle ACB=75^{\circ}$.

your answer is correct !,will you try to use geometry ?
(let geometry also be your close friend)

 

[anemone]

your answer is correct !,will you try to use geometry ?
(let geometry also be your close friend)

I in fact tried to tackle it geometrically and I thought it must have everything to do to prove that

Spoiler

the quadrilateral $AGDC$ is cyclic

but all of my attempts had been proven to be exercises in futility.(Doh)

 

[Albert1]

$\triangle ABC ,\angle ABC =45^o,\,\,point \,\, D \,\, on \,\, \overline{BC}, \,\,and :$
$2\overline {BD}=\overline {CD},\,\,\angle DAB=15^o$
please find :$\angle ACB=?$

My solution:

Spoiler

View attachment 3573