Question

Află măsurile unghiurilor AOB, BOC, COD, DOA în jurul punctului O cu proprietatea 3 m(∡AOB) = 4 m(∡COD), 4 m(∡AOC) = 5 m(∡BOD) și m(∡AOD) = 7 m(∡BOC).

Mulțumesc!

Answer 1

Voi nota cu:

$\measuredangle X$

masura unghiului X.

In primul rand, deoarece cele 4 unghiuri sunt unghiuri in jurul unui punct, putem scrie ca:

$\measuredangle AOB+\measuredangle BOC+\measuredangle COD+\measuredangle DOA=360^{\circ}$

Pentru a fi mai usor, voi nota masura unghiului AOB cu x.

$3\cdot\measuredangle AOB=4\cdot\measuredangle COD\implies \measuredangle COD=3/4\cdot\measuredangle AOB =3/4\cdot x$

Scriem unghiul AOC ca:

AOC = AOB + BOC

$\measuredangle AOC =\measuredangle AOB+\measuredangle BOC=x+\measuredangle BOC$

Scriem unghiul BOD ca:

BOD = BOC + COD

$\measuredangle BOD = \measuredangle BOC + \measuredangle COD = \measuredangle BOC+ 3/4\cdot x$

Dar stim ca:

$4\cdot\measuredangle AOC=5\cdot\measuredangle BOD$

adica:

$4(x+\measuredangle BOC)=5(\measuredangle BOC+3/4\cdot x)\implies$

$4\cdot x+4\cdot\measuredangle BOC=5\cdot\measuredangle BOC + 15/4\cdot x$

$\measuredangle BOC = 4x-15/4\cdot x =1/4\cdot x$

Apoi:

$\measuredangle DOA=\measuredangle AOD =7\cdot\measuredangle BOC=7/4\cdot x$

Introducem acum tot ce am obtinut in prima ecuatie:

$x+1/4\cdot x+3/4\cdot x+7/4\cdot x=360^{\circ}\implies$

$15/4\cdot x =360^{\circ}\implies x=4/15\cdot360^{\circ}=96^{\circ}$

Asadar:

$\measuredangle AOB=x=96^{\circ}$

$\measuredangle BOC=1/4\cdot x=24^{\circ}$

$\measuredangle COD=3/4\cdot x=72^{\circ}$

$\measuredangle DOA =7/4\cdot x =168^{\circ}$