In triangle abc angle abc is equal to 80 degree angle ACB is equals to 40 degree ap is the bisector of angle BAC and aq is perpendicular to BC find angle p a q

Given in the problem:

- \(\angle ABC = 80^\circ\)
- \(\angle ACB = 40^\circ\)
- \(AP\) is the bisector of \(\angle BAC\)
- \(AQ\) is perpendicular to \(BC\)
- We need to find \(\angle PAQ\).

### Step 1: Find \(\angle BAC\)
In \(\triangle ABC\), the sum of all angles is \(180^\circ\). So:
\[
\angle BAC = 180^\circ - \angle ABC - \angle ACB
\]
\[
\angle BAC = 180^\circ - 80^\circ - 40^\circ = 60^\circ
\]

### Step 2: Find \(\angle BAP\) and \(\angle CAP\)
Since \(AP\) is the angle bisector of \(\angle BAC\):
\[
\angle BAP = \angle CAP = \frac{1}{2} \times \angle BAC = \frac{60^\circ}{2} = 30^\circ
\]

### Step 3: Consider the triangle \(PAQ\)
\(AQ\) is perpendicular to \(BC\), so \(\angle AQC = \angle AQB = 90^\circ\). Now, we need to find \(\angle PAQ\).

Since \(AQ\) is perpendicular to \(BC\) and \(AP\) is the angle bisector, \(AP\) and \(AQ\) are in the same plane with \(\angle BAC = 60^\circ\). So the required angle \(\angle PAQ\) is formed by the bisector and the perpendicular.

### Step 4: Calculate \(\angle PAQ\)
The angle between the angle bisector and the perpendicular from the vertex to the opposite side in an isosceles triangle (where the vertex angle is \(60^\circ\)) can be calculated as:

\[
\angle PAQ = 90^\circ - \frac{\angle BAC}{2} = 90^\circ - 30^\circ = 60^\circ
\]

### Final Answer:
\(\angle PAQ = 60^\circ\).

Would you like any further explanation or details about any steps?

Here are some additional questions related to this concept:
1. How would the solution change if the angle \( \angle ABC \) was \(70^\circ\) instead of \(80^\circ\)?
2. Can you find the length of \(AP\) or \(AQ\) given the sides of the triangle?
3. How does the location of point \(P\) affect the size of angle \(PAQ\)?
4. What would \(\angle PAQ\) be if \(AP\) was the median instead of the angle bisector?
5. How can we generalize the relationship between the angle bisector, perpendicular, and angles in any triangle?

**Tip**: In any triangle, the angle bisector theorem and the properties of perpendiculars can provide valuable insight into solving complex geometric problems.

Explore how to find angle PAQ in triangle ABC using the angle bisector theorem and perpendicular properties. Given angles ABC = 80°, ACB = 40°, AP is the angle bisector of angle BAC, and AQ is perpendicular to BC. Detailed step-by-step solution provided.

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