Question

Na figura, DA = DB = DC. Qual a medida do ângulo ß?

Answer 1

Resposta:

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Explicação passo-a-passo:

Answer 2

!! Veja a imagem !!

$\underline{\Delta\text{ABD} : \text{is{\'o}sceles} }\\\\ \text{AD}=\text{DB} \\\\ \angle\text{A} = \angle\text{B}=25^\circ \\\\ \text{Da{\'i} pelo teoerma do {\^a}ngulo externo o {\^a}ngulo D do} \ \Delta\text{BDC} \ \text{vale} : \\\\ \angle\text D = \angle\text A+\angle \text B \\\\ \angle \text D = 25^\circ+25^\circ \\\\ \angle\text D = 50^\circ \\\\\\ \Delta_\text{DBC} \ \text{is{\'o}sceles DB = DC , logo : } \\\\ \angle\text B =\angle \text C$

$\\\\ \angle\text D +\angle \text B +\angle\text C =180^\circ \\\\ 2\angle\text B +50^\circ=180^\circ \\\\ 2\angle\text B= 130^\circ \\\\ \angle \text B = 65^\circ = \angle \text C \\\\\\ \text{Ent{\^a}o temos que}:\\\\\ 25^\circ+65^\circ+\beta = 180^\circ \\\\ \huge\boxed{\beta = 90^\circ} \checkmark$