In `Delta ABC`, P is an interior point such that `angle PAB = 10^(@), anglePBA = 20^(@), anglePCA = 30^(@), anglePAC = 40^(@) " then " DeltaABC` is

Options

A. isosceles
B. right angled
C. equilateral
D. obtuse angled

Explanation

To solve the problem, we need to analyze the angles given in triangle ABC with point P inside it. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Angles
We are given:
- \( \angle PAB = 10^\circ \)
- \( \angle PBA = 20^\circ \)
- \( \angle PCA = 30^\circ \)
- \( \angle PAC = 40^\circ \)

We need to find the angles of triangle ABC.

### Step 2: Determine Angle PBC
Let \( \angle PBC = x \). We can now express the angles at point B:
- \( \angle ABC = \angle PBA + \angle PBC = 20^\circ + x \)

### Step 3: Calculate Angle ACB
Using the angles around point P, we can find \( \angle ACB \):
- \( \angle ACB = \angle PCA + \angle PAC = 30^\circ + 40^\circ = 70^\circ \)

### Step 4: Use Angle Sum Property in Triangle ABC
The sum of the angles in triangle ABC must equal \( 180^\circ \):
\[ \angle ABC + \angle ACB + \angle BAC = 180^\circ \]
Substituting the known values:
\[ (20^\circ + x) + 70^\circ + \angle BAC = 180^\circ \]
This simplifies to:
\[ 90^\circ + x + \angle BAC = 180^\circ \]
Thus:
\[ \angle BAC = 90^\circ - x \]

### Step 5: Find the Value of x
Now we have:
- \( \angle ABC = 20^\circ + x \)
- \( \angle ACB = 70^\circ \)
- \( \angle BAC = 90^\circ - x \)

Using the sine law as mentioned in the transcript, we set up the equation:
\[ \sin(20^\circ) \cdot \sin(80^\circ - x) \cdot \sin(40^\circ) = \sin(30^\circ) \cdot \sin(10^\circ) \cdot \sin(x) \]

### Step 6: Simplify and Solve
Using the sine double angle identity:
\[ \sin(20^\circ) = 2 \sin(10^\circ) \cos(10^\circ) \]
We can substitute this into our equation and simplify. After simplification, we find that \( x = 60^\circ \).

### Step 7: Find Angles of Triangle ABC
Now substituting \( x = 60^\circ \):
- \( \angle ABC = 20^\circ + 60^\circ = 80^\circ \)
- \( \angle BAC = 90^\circ - 60^\circ = 30^\circ \)
- \( \angle ACB = 70^\circ \)

### Step 8: Identify Triangle Type
From the angles:
- \( \angle A = 30^\circ \)
- \( \angle B = 80^\circ \)
- \( \angle C = 70^\circ \)

Since \( \angle A \neq \angle B \neq \angle C \), triangle ABC is not equilateral or isosceles. The angles indicate that it is an acute triangle.

### Conclusion
Thus, triangle ABC is an acute triangle.