Question
ABC is an isosceles triangle and AC, BC are the tangents at M and N respectively. DE is the diameter of the circle. ∠ADP = ∠BEQ = 100∘. What is value of ∠PRD?
Answer: Option C
:
C
ADB is a straight line. Bylinear pair axiom,
∠ADP + ∠PDB =180∘
100 + ∠PDB =180∘
∠PDB =80∘
Similarly ∠QED =80∘
We have, ∠DPE = 90∘(angle subtended by a diameter)
In △DPE,
∠DPE + ∠PED + ∠EDP = 180
[Angle sum property of a triangle]
∠PED = 10∘
Similarly ∠QDE =10∘
In △DRE,
∠DRE + ∠RDE + ∠RED = 180
[Angle sum property of a triangle]
∠DRE = 160∘
PRE is a straight line.Bylinear pair axiom,
∠PRD + ∠DRE =180∘
∠PRD + 160 =180∘
∠PRD =20∘
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:
C
ADB is a straight line. Bylinear pair axiom,
∠ADP + ∠PDB =180∘
100 + ∠PDB =180∘
∠PDB =80∘
Similarly ∠QED =80∘
We have, ∠DPE = 90∘(angle subtended by a diameter)
In △DPE,
∠DPE + ∠PED + ∠EDP = 180
[Angle sum property of a triangle]
∠PED = 10∘
Similarly ∠QDE =10∘
In △DRE,
∠DRE + ∠RDE + ∠RED = 180
[Angle sum property of a triangle]
∠DRE = 160∘
PRE is a straight line.Bylinear pair axiom,
∠PRD + ∠DRE =180∘
∠PRD + 160 =180∘
∠PRD =20∘
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