In triangle $$ ABC $$, altitudes $$ AD $$ and $$ BE $$ are drawn to the corresponding bases. If $$ \angle BAC = 45^\circ $$ and $$ \angle ABC = heta $$, then
$$ \frac{AD}{BE} $$ equals:

Question

In triangle $$ ABC $$, altitudes $$ AD $$ and $$ BE $$ are drawn to the corresponding bases. If $$ \angle BAC = 45^\circ $$ and $$ \angle ABC = 	heta $$, then  
$$ \frac{AD}{BE} $$ equals:

Accepted Answer

To solve this problem:

1. **Define the Sides of the Triangle**:  
   Let the sides opposite to vertices $$ A, B, C $$ be $$ a, b, c $$, respectively.

2. **Area of the Triangle Using Altitudes**:  
   The area of triangle $$ ABC $$, denoted as $$ \Delta $$, can be expressed in two ways:  
   - Using altitude $$ AD $$:  
     $$
     \Delta = \frac{1}{2} a \cdot AD
     $$
   - Using altitude $$ BE $$:  
     $$
     \Delta = \frac{1}{2} b \cdot BE
     $$

3. **Area Using Sine Rule**:  
   The area can also be written using the sine rule:  
   - Using angle $$ B $$:  
     $$
     \Delta = \frac{1}{2} a c \sin(B)
     $$
   - Using angle $$ A = 45^\circ $$:  
     $$
     \Delta = \frac{1}{2} b c \sin(A)
     $$

4. **Relate Sides Using Sine Rule**:  
   From the sine rule:  
   $$
   \frac{a}{b} = \frac{\sin(B)}{\sin(A)} = \frac{\sin(	heta)}{\sin(45^\circ)}
   $$
   Since $$ \sin(45^\circ) = 1/\sqrt{2} $$:  
   $$
   a/b = \sqrt{2} \sin(	heta)
   $$

5. **Relate Altitudes**:  
   From the expressions for area:  
   $$
   a/AD = b/BE
   $$
   Thus:  
   $$
   AD/BE = a/b = \sqrt{2} \sin(	heta)
   $$

Hence, the correct answer is **$$ A. \sqrt{2} \sin(	heta) $$**.

Answer

Answer

Answer