Find angle of this polygon

Question

I apologize for this simple question but unfortunately I'm not good at geometry. Consider a polygon in the following image

we have $$AB=BC=AD.$$ I want to find he angle $C$

How can I find the angle? Can anybody help me? .

Answer 1

Create a second triangle that is congruent to $\triangle ABD$

Use what you know about isosceles triangles to find $\angle ABD, \angle DBA'$ and $\angle A'BC$

From this we find $\angle BA'C$ and $\angle A'CD$

Answer 2

We know that $\bbox[lightgreen]{AB=BC=AD}$, $\implies$ $\triangle{ABD}$ is isosceles, $\implies$ $\angle{ABD}=\angle{ADB}=\frac{180^{0}-74^{0}}{2}=53^{0}$, $\implies$ $\angle{DBC}=113^{0}$.

Let to be $\bbox[lightblue]{a=AB=BC=AD}$.

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Using the well very known law of sines relatively $\triangle{ADB}$ we can get that: $$ \frac{DB}{\sin(74^{0})}=\frac{a}{\sin(53^{0})}\implies{DB}=\frac{a\cdot{\sin(74^{0})}}{\sin(53^{0})}. $$

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Using the well very known law of cosines relatively $\triangle{DBC}$ we can get that: $$ DC^{2}=DB^{2}+BC^{2}-2\cdot{DB}\cdot{BC}\cdot\cos(\angle{DBC})= $$ $$ =\frac{a^{2}\sin^{2}(74^{0})+a^{2}\sin^{2}(53^{0})-2a^{2}\sin(74^{0})\cos(113^{0})\sin(53^{0})}{\sin^{2}(53^{0})}= $$ $$ =\frac{a^{2}(\sin^{2}(74^{0})+\sin^{2}(53^{0})-\sin(106^{0})\sin(74^{0}))}{\sin^{2}(53^{0})},\implies \\ \implies{DC}=\frac{a}{\sin(53^{0})}\left(\sin^{2}(74^{0})+\sin^{2}(53^{0})-\sin(106^{0})\sin(74^{0})\right)^{0.5}. $$

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Using the well very known law of sines relatively $\triangle{DBC}$ we can get that: $$ \frac{DB}{\sin(\angle{BCD})}=\frac{DC}{\sin(\angle{DBC})}\implies\sin(\angle{BCD})=\frac{DB\cdot\sin(\angle{DBC})}{DC}= \\ =\frac{a\cdot\sin(74^{0})\sin(113^{0})\sin(53^{0})}{a\sin(53^{0})\left(\sin^{2}(74^{0})+\sin^{2}(53^{0})-\sin(106^{0})\sin(74^{0})\right)^{0.5}}.\implies $$ $$ \implies \bbox[pink] { \angle{BCD}=\arcsin\left(\frac{\sin(74^{0})\sin(113^{0})}{\left(\sin^{2}(74^{0})+\sin^{2}(53^{0})-\sin(106^{0})\sin(74^{0})\right)^{0.5}}\right). } $$

Good luck!

Answer 3

$ABD$ is an isosceles triangle $→$ $\angle ABD= \angle ADB =53°$
Draw two circles centers at $A$ and $B$ with radius $AB=BC$
Triangle $ABE$ is an equilateral triangle $→$ $\angle BAE=60°$
Central angle $\angle BAE=60° → \angle BDE=30°$
In triangle $BCD$, the angle $\angle BCD=37°$

Answer 4

Another answer:

Construct an equilateral triangle ADE.

By simple angle chasing you can find some angles in the picture below

Now you know that $AE=AB$ and $\angle EAC= \angle BAC=7$ degrees

Then you can find that ABCE is a rhombus

Then that all side lengths are equal.

You already know that $\angle AED=60, \angle AEC=166$ (Since ABCE is a rhombus)

Then you can find $\angle AEC$ as follows

$\angle AEC+60+166=360$

$\angle AEC=134$

Since $\triangle DEC$ is isosceles $\angle ECD=23$

Finally $\angle BCD=23+7+7$

$\angle BCD=37$ degrees