Exterior Angle Theorem Application
Consider $\Delta PQR$, where QR is extended to S, forming a straight line QRS. The angle $\angle PRS = 110^\circ$ is the exterior angle to $\Delta PQR$ at vertex R.
The exterior angle theorem states that an exterior angle equals the sum of the two opposite interior angles ($\angle PQR$ and $\angle RPQ$).
Calculating $\angle PQR$
Using the exterior angle theorem:
$ \angle PRS = \angle PQR + \angle RPQ $
Substitute the given values $\angle PRS = 110^\circ$ and $\angle RPQ = 55^\circ$:
$ 110^\circ = \angle PQR + 55^\circ $
Solve for $\angle PQR$:
$ \angle PQR = 110^\circ - 55^\circ $
$ \angle PQR = 55^\circ $
Final Result for $\angle PQS$
Since Q, R, S are collinear, $\angle PQS$ is the same angle as $\angle PQR$.
Therefore, $\angle PQS = 55^\circ$.
Similar Questions
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Important Questions from Triangles
What is the circumcenter of the triangle ABC?
What is the centroid of the triangle ABC?
What is the foot of the altitude from the vertex A of the triangle ABC?
In ΔABC, D is a point on BC such that ∠ADB = 2∠DAC, ∠BAC = 70° and ∠B = 56°. What is the measure of ∠ADC?
In ΔABC, ∠A = 66° and ∠B = 50 °. If the bisectors of ∠B and ∠C meet at P, then ∠BPC – ∠PCA = ?
