In a` triangleABC`, CP and BQ intersect each other at R inside the triangle, where P and Q lie on the sides AB and AC respectively. `angleBAC=50^@, angleABC=70^@, angleACB=angleAPQ` and PC=BC. Find `anglePBQ.`

Options

A. `10^@`
B. `15^@`
C. `20^@`
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Draw the Triangle
Draw triangle ABC with angles:
- \( \angle BAC = 50^\circ \)
- \( \angle ABC = 70^\circ \)
- \( \angle ACB = 180^\circ - (50^\circ + 70^\circ) = 60^\circ \)

### Step 2: Identify Points P and Q
Place point P on side AB and point Q on side AC. Draw lines CP and BQ intersecting at point R inside the triangle.

### Step 3: Define Angles
Let \( \angle APQ = \angle ACB = 60^\circ \) (as given in the problem). Also, since \( PC = BC \), we have two equal sides.

### Step 4: Use the Isosceles Triangle Property
Since \( PC = BC \), the angles opposite these sides must be equal:
- \( \angle PBC = \angle ACB = 60^\circ \)
- \( \angle BPC = 70^\circ \) (since \( \angle ABC = 70^\circ \))

### Step 5: Find \( \angle PBQ \)
Now, we need to find \( \angle PBQ \). We know:
- \( \angle PBC = 60^\circ \)
- \( \angle ABC = 70^\circ \)

Using the angle sum property in triangle PBC:
\[ \angle PBC + \angle BPC + \angle PBQ = 180^\circ \]
Substituting the known angles:
\[ 60^\circ + 70^\circ + \angle PBQ = 180^\circ \]
\[ 130^\circ + \angle PBQ = 180^\circ \]
\[ \angle PBQ = 180^\circ - 130^\circ = 50^\circ \]

### Final Answer
Thus, the measure of \( \angle PBQ \) is \( 50^\circ \).

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