We know the sum of all angles in a triangle is
![180^( \circ)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/xq7ukt6bor2aneicrus0sdjfjn8mlp34y3.png)
and they have given both the triangles two of the angles.
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For the one on the right, we have
![53+80=133](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ywgydh3ccifno53436ku726w2rx2t7e7gd.png)
![180-133=47](https://img.qammunity.org/2019/formulas/mathematics/middle-school/xf09w0jspkvntxv491z3j38t4powvvz6qc.png)
The bottom left angle of the triangle on the right is
![47^( \circ)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/eic75u58uck8qxddksmj96ey9sgivio13g.png)
.
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For the triangle on the left, we do the same process.
The bottom right angle of the triangle on the left is
![68^( \circ)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/kr303heegc8xppbpbuz6uu44t300y76gru.png)
.
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Now we can find the last angle of the smaller triangle created by the overlapping larger triangles.
Using the same method as the previous ones, we get
![65^( \circ)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/okr5km3cmub47kkkq4005r23aj1u8350rb.png)
.
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A straight angle is
![180^( \circ)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/xq7ukt6bor2aneicrus0sdjfjn8mlp34y3.png)
.
![180-65=\boxed{115}](https://img.qammunity.org/2019/formulas/mathematics/middle-school/fjnu9kvaaynxcm2dml8htyhjje822watsg.png)
![x=115](https://img.qammunity.org/2019/formulas/mathematics/middle-school/f9n60yoxe5qjtzf52vwq4940ejniuo5zl9.png)