Question

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Answer 1

Doua drepte concurente formeaza 4 unghiuri cu masuri egale doua cate doua: masurile unghiurilor opuse la varf.

Astfel ca:

$m(\measuredangle A_1)=m(\measuredangle A_3)$

$m(\measuredangle A_2)=m(\measuredangle A_4)$

Suma tuturor celor 4 unghiuri este de 360°, pentru ca atatea grade are o rotatie completa in jurul punctului de concurenta a dreptelor.
Mai mult decat atat, suma a 2 unghiuri consecutive (dintre cele 4) este de 180°.
Adica:

$m(\measuredangle A_1)+m(\measuredangle A_2)=180^\circ$

$m(\measuredangle A_3)+m(\measuredangle A_4)=180^\circ$


a. $m(\measuredangle A_1)=25^\circ\implies m(\measuredangle A_3)=25^\circ$

$m(\measuredangle A_2)=180^\circ-m(\measuredangle A_1)=180^\circ-25^\circ=155^\circ$

$\implies m(\measuredangle A_4)=155^\circ$

b. $m(\measuredangle A_1)=37^\circ43'\implies m(\measuredangle A_3)=37^\circ43'$

$m(\measuredangle A_2)=180^\circ-37^\circ43'$

Stim ca $1^\circ=60'$

Deci

$m(\measuredangle A_2)=179^\circ-37^\circ+1^\circ-43'=142^\circ+60'-43'=142^\circ17'$

$\implies m(\measuredangle A_4)=142^\circ17'$

c. $m(\measuredangle A_1)=92^\circ13'\implies m(\measuredangle A_3)=92^\circ13'$

$m(\measuredangle A_2)=180^\circ-92^\circ13'=179^\circ-92^\circ+60'-13'=87^\circ47'$

$\implies m(\measuredangle A_4)=87^\circ47'$