Question

1. In Figure 3, RK is the angle bisector of ORC.
a. Name the postulate/theorem that justifies the congruency of AROK and ARCK.
b. If mZORC = 50, what is mZRKC?
If RO = 3x - 12 and RC = 2(x - 2), find x and RO.
C.​

Answer 1

Solution :-

In ∆ROK and ∆RCK , we have,

→ ∠ORK = ∠CRK { RK is angle bisector of ∠ORC.}

→ ∠KOR = ∠KCR { By construction = 90°.}

so,

→ ∆ROK ∆RCK { By AA postulate.}

then,

→ RO / RC = RK / RK

→ RO = RC

→ 3x - 12 = 2(x - 2)

→ 3x - 12 = 2x - 4

→ 3x - 2x = 12 - 4

→ x = 8 .

therefore,

→ RO = 3x - 12 = 3*8 - 12 = 24 - 12 = 12 .

hence,

→ ∠RKC = 180° - (90° + 50°/2) = 180° - (90° + 25°) = 180° - 115° = 65° .

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