Given: Circle O and Circle Q intersect at R and S;
mRVS=60; mRUS=120
Prove: OR is tangent to circle Q
QR is tangent to circle O

Solution:
Given that circle O and circle Q intersect at R and S, and that mRVS = 60° and mRUS = 120°.
To prove that OR is tangent to circle Q, we need to show that mROS = 90°.
Since OQ is a straight line, mROS + mRQS = 180°. Given that mRVS = 60°, mRQS = 60° as well. Therefore, mROS = 180° - mRQS - mRVS = 180° - 60° - 60° = 60°.
Since mROS = 60°, we can conclude that OR is tangent to circle Q.
The same reasoning can be applied to prove that QR is tangent to circle O, using similar steps.

Answered by qwarrow | 2024-06-24

To prove (OR) tangent to (Q): m ∠ OQR = 3 0 ∘ , m ∠ ORQ = 9 0 ∘ , thus OR is tangent.
To prove (QR) tangent to (O): m ∠ RQS = 6 0 ∘ , m ∠ QRO = 3 0 ∘ , hence (QR) is tangent.
To prove that ( OR ) is tangent to circle ( Q ) and ( QR ) is tangent to circle ( O ), we will use the properties of tangents and the inscribed angles theorem.
Let's start with proving ( OR ) is tangent to circle ( Q ):
Given: Circle ( O ) and Circle ( Q ) intersect at ( R ) and ( S ).
∠ R V S = 6 0 ∘ and ∠ R U S = 12 0 ∘
To prove: ( OR ) is tangent to circle ( Q )
Proof:

Claim : ∠ OQR = 3 0 ∘


- Reasoning: Since ∠ R V S = 6 0 ∘ , it's an inscribed angle subtended by the arc ( RS ). By the inscribed angle theorem, ∠ RQS = 6 0 ∘ .[/tex]Since ( QR ) is a tangent,[tex] ∠ OQR is half of ∠ RQS , so ∠ OQR = 2 1 ​ ⋅ 6 0 ∘ = 3 0 ∘ .

Claim : ∠ ORQ = 9 0 ∘


- Reasoning: In a circle, a radius drawn to a tangent forms a right angle. ( OR ) is a radius of circle ( Q ), and ( QR ) is a tangent to circle ( Q ), hence ∠ ORQ = 9 0 ∘ .[/tex]

Conclusion : ( OR ) is tangent to circle ( Q ) (By definition of a tangent).

Now, let's prove that ( QR ) is tangent to circle ( O ):
Given: Circle ( O ) and Circle ( Q ) intersect at ( R ) and ( S ).
[ ∠ R V S = 6 0 ∘ and ∠ R U S = 12 0 ∘
To prove: ( QR ) is tangent to circle ( O )
Proof:

Claim : ∠ RQS = 6 0 ∘


- Reasoning: Given that ∠ R V S = 6 0 ∘ , ∠ RQS is an inscribed angle subtended by the arc ( RS ). Hence, ∠ RQS = 6 0 ∘ .]

Claim : [tex] ∠ QRO = 3 0 ∘


- Reasoning: Since \angle RUS = 120^\circ \), \( \angle RQS = 60^\circ \), and \( RU \) is a tangent, \( \angle QRO \) is half of \( \angle RQS \). So, \( \angle QRO = \frac{1}{2} \cdot 60^\circ = 30^\circ \

Conclusion : ( QR ) is tangent to circle ( O ) (By definition of a tangent).

Therefore, we have proved that ( OR ) is tangent to circle ( Q ) and ( QR ) is tangent to circle ( O ).

Answered by susmthaIRA | 2024-06-25

By calculating angles around the intersection points of circles O and Q, we prove that the angles formed support the notion that OR and QR are tangent lines. Specifically, we check the angle measures to confirm that each tangent meets the respective circle at a right angle. Thus, both line segments OR and QR are tangents to their respective circles.
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Answered by qwarrow | 2024-12-02