Math Problem Statement

Solve it step by step

In a triangle PQR, vertical angle P = 50°. The circle passing through Q and R intersects PQ in S and PR in T. ST is joined. Find angle PST.

Solution

Let's solve the problem step by step. The given diagram and the question indicate that:

  1. PQR\triangle PQR is isosceles with P=50\angle P = 50^\circ.
  2. A circle passes through points QQ and RR, intersecting PQPQ at SS and PRPR at TT.
  3. STST is joined, and we need to find PST\angle PST.

Step 1: Identify key angles in PQR\triangle PQR

  • Since PQR\triangle PQR is isosceles with PQ=PRPQ = PR, the base angles are equal. Let these angles be xx.
  • The sum of angles in PQR\triangle PQR is 180180^\circ: P+Q+R=180\angle P + \angle Q + \angle R = 180^\circ Substituting P=50\angle P = 50^\circ and Q=R=x\angle Q = \angle R = x: 50+x+x=1802x=130x=6550^\circ + x + x = 180^\circ \quad \Rightarrow \quad 2x = 130^\circ \quad \Rightarrow \quad x = 65^\circ Therefore, Q=R=65\angle Q = \angle R = 65^\circ.

Step 2: Use properties of the circle

  • PSQ\angle PSQ and PTR\angle PTR are angles subtended by the chord STST at the points SS and TT.
  • By the cyclic quadrilateral property, opposite angles in a cyclic quadrilateral add up to 180180^\circ.
  • The circle is symmetric about PQR\triangle PQR. Thus: PSQ=PTS\angle PSQ = \angle PTS

Step 3: Compute PST\angle PST

  • Since STST is joined, PST\angle PST subtends the arc QSQS at point PP.
  • The sum of angles subtended by the same arc in a cyclic quadrilateral equals the angle at the center or the opposite segment.

Using symmetry:

Do you continue Questions.,Expans. Let's complete the solution step-by-step!


Step 3: Compute PST\angle PST (continued)

  • From the geometry, PST\angle PST is the external angle at PP subtended by chord STST.
  • Using properties of cyclic quadrilaterals and the fact that angles on the same arc are equal, PST\angle PST is equal to QTR\angle QTR, which subtends the same arc QRQR.

Now, we know:

  • QPR=50\angle QPR = 50^\circ, and PQR\triangle PQR is isosceles.
  • By angle properties: [ \angle PST = 90^\circ - \frac{\angle P}{2} = \textcal

EXTRA

Let’s complete the calculation:

In the problem:

  • PST\angle PST subtends the arc QRQR, and we know that PQR\triangle PQR is isosceles.
  • The angle subtended by chord QRQR at the center is 18050=130 180^\circ - 50^\circ = 130^\circ. Now to finalize:

Since PQR\triangle PQR is symmetric, we focus on the angles subtended in the circle:

  1. The key angle subtended by the chord STST at PP can be derived directly from the geometry.
  2. Using symmetry and cyclic quadrilateral properties, PST=65\angle PST = \boxed{65^\circ}, which matches the base angle Q\angle Q.

Do you want further clarification?

Here are five related questions to explore:

  1. How are cyclic quadrilateral properties used to simplify such problems?
  2. What are the specific properties of an isosceles triangle relevant to this problem?
  3. How does symmetry in circles affect angle relationships?
  4. How can we apply external angle properties in other cyclic quadrilateral problems?
  5. Can we solve this using alternative methods like trigonometry?

Tip: Always look for cyclic quadrilateral rules in geometry problems involving circles—they often simplify complex relationships.

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Math Problem Analysis

Mathematical Concepts

Triangles
Circles
Cyclic Quadrilaterals
Isosceles Triangle Properties
Angle Sum Property

Formulas

Sum of angles in a triangle: angle P + angle Q + angle R = 180°
Cyclic quadrilateral property: opposite angles sum to 180°

Theorems

Isosceles triangle theorem
Properties of angles in a cyclic quadrilateral

Suitable Grade Level

Grades 9-12