PQ and PR are two tangents drawn from an external point P.
∴ PQ = PR
[ the lengths of tangents drawn from an external point to a circle are equal]
⇒∠PQR=∠QRP [ angles opposite to equal sides are equal]
Now, in
Δ PQR
∠PQR+∠QRP+∠RPQ=180∘ [ sum of all interior angles of any triangle is
180∘]
⇒ ∠PQR +
∠PQR +
30∘=180∘ ⇒2∠PQR=180∘−30∘ ⇒∠PQR=180∘−30∘2=75∘ Since SR
∥ QP
∴∠SRQ=∠RQP=75∘ [alternate interior angles]
Also,
∠PQR=∠QSR=75∘ [ by alternate segment theorem]
In
Δ QRS
∠Q+∠R+∠S=180∘ [ sum of all interior angles of any triangle is
180∘]
⇒∠Q=180∘−(75∘+75∘) =30∘ ∠RQS=30∘