Triangle espacate

DBM is isosceles and the external angle $\angle ABM$ is divided equally between $\angle BDM$ and $\angle BMD$ . Also, A, B and D are collinear, so are A, C and E, and $BD=BM=CM=CE$ , so DE is parallel to BC. Hence, $\angle BMD = \angle MDE = \frac { \angle ABM }{ 2 }$ . The same goes for $\angle DEM$ . As a result, the congruent angles ( ${ \alpha }_{ n }$ )of the resulting isosceles triangle is always half of those of the former ones ( ${ \alpha }_{ n-1 }$ ). Let $\angle ABM$ be ${ \alpha }_{ 0 }$ . We know that $\angle BAC = 20º$ . Thus:

${ \alpha }_{ n } = \frac { { \alpha }_{ n-1 } }{ 2 }$

${ \alpha }_{ 0 } = 80º$

${ \alpha }_{ n } = { \alpha }_{ 0 }\cdot{ \left( \frac { 1 }{ 2 } \right) }^{ n }$

${ \alpha }_{ n }=80º\cdot { 2 }^{ -n }$

Let $\angle BAC$ be ${ \theta }_{ 0 }$ , $\angle DME$ be ${ \theta }_{ 1 }$ and so on. Therefore, we have that:

${ \theta }_{ n } = 180º - 2\cdot { \alpha }_{ n }$

${ \theta }_{ n } = 180º - 2\cdot80º\cdot { 2 }^{ -n }$

For ${ \theta }_{ n } = 175º$ , it goes like this:

$175º = 180º - 2\cdot80º\cdot { 2 }^{ -n }$

$160º\cdot { 2 }^{ -n } = 5º$

${ 2 }^{ -n } = \frac { 1 }{ 32 } = { 2 }^{ -5 }$

$n=5$

Hence, we need to form 5 triangles in order to get to the angle of 175º.