R is circumradii of `DeltaABC, H` is orthocentre, `R_(1), R_(2), R_(3)` are circumradii of `Delta AHB, Delta BHC.` If AH produced meet the circumradii of ABC at M and intersect BC at L,
`angle AHB =180^(@)-C`
`(c )/(sin (180^(@)-C))=2R_(1)`
`(c)/(sin C) =2R_(1)`
`R_(1)=R`
Area of `Delta AHB`

Options

A. a. 2R cos A cos B cos C
B. b. `R^(2)` cos A cos B cos C
C. c. `2R^(2)` cos A cos B sin C
D. d. None of the above

Answer

To find the area of triangle \( AHB \), we can follow these steps:

### Step 1: Understand the Geometry
We have triangle \( ABC \) with circumradius \( R \) and orthocenter \( H \). The circumradii of triangles \( AHB \), \( BHC \), and \( CHA \) are denoted as \( R_1 \), \( R_2 \), and \( R_3 \) respectively. We know that \( \angle AHB = 180^\circ - C \).

### Step 2: Use the Relationship of Circumradii
From the properties of triangles, we have:
\[ \frac{c}{\sin(180^\circ - C)} = 2R_1 \]
Since \( \sin(180^\circ - C) = \sin C \), we can write:
\[ \frac{c}{\sin C} = 2R_1 \]
Thus, we find:
\[ R_1 = R \]
This means that the circumradius of triangle \( AHB \) is equal to the circumradius of triangle \( ABC \).

### Step 3: Find the Lengths \( AH \) and \( BH \)
Using the right triangle properties, we can express \( AH \) and \( BH \) in terms of the circumradius \( R \):
1. For \( AH \):
\[ AH = 2R \cos A \]
2. For \( BH \):
\[ BH = 2R \cos B \]

### Step 4: Calculate the Area of Triangle \( AHB \)
The area \( A \) of triangle \( AHB \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times AH \times BH \times \sin(\angle AHB) \]
Substituting the values we found:
\[ \text{Area} = \frac{1}{2} \times (2R \cos A) \times (2R \cos B) \times \sin(180^\circ - C) \]
Since \( \sin(180^\circ - C) = \sin C \), we have:
\[ \text{Area} = \frac{1}{2} \times 4R^2 \cos A \cos B \sin C \]
This simplifies to:
\[ \text{Area} = 2R^2 \cos A \cos B \sin C \]

### Final Result
Thus, the area of triangle \( AHB \) is:
\[ \text{Area} = 2R^2 \cos A \cos B \sin C \]

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