The difference between two angles of a triangle whose magnitude is in the ratio 10:7 is `20^(@)` less than the third angle. The third angle is:

Options

A. `80^(@)`
B. `56^(@)`
C. `44^(@)`
D. `70^(@)`

Answer

To solve the problem step by step, we will use the information given about the angles of the triangle.

### Step 1: Define the Angles
Let the three angles of the triangle be \( A \), \( B \), and \( C \). According to the problem, the ratio of the two angles \( A \) and \( B \) is \( 10:7 \).

### Step 2: Express Angles in Terms of a Variable
Since \( A \) and \( B \) are in the ratio \( 10:7 \), we can express them as:
\[ A = 10x \quad \text{and} \quad B = 7x \]
where \( x \) is a common multiplier.

### Step 3: Use the Triangle Angle Sum Property
The sum of the angles in a triangle is \( 180^\circ \). Therefore, we have:
\[ A + B + C = 180^\circ \]
Substituting the expressions for \( A \) and \( B \):
\[ 10x + 7x + C = 180^\circ \]
This simplifies to:
\[ 17x + C = 180^\circ \]

### Step 4: Express the Difference Between the Angles
The problem states that the difference between the two angles \( A \) and \( B \) is \( 20^\circ \) less than the third angle \( C \). Thus, we can write:
\[ A - B = C - 20^\circ \]
Substituting the expressions for \( A \) and \( B \):
\[ 10x - 7x = C - 20^\circ \]
This simplifies to:
\[ 3x = C - 20^\circ \]
Rearranging gives us:
\[ C = 3x + 20^\circ \]

### Step 5: Substitute \( C \) in the Angle Sum Equation
Now we have two equations involving \( C \):
1. \( C = 3x + 20^\circ \)
2. \( 17x + C = 180^\circ \)

Substituting the first equation into the second:
\[ 17x + (3x + 20^\circ) = 180^\circ \]
This simplifies to:
\[ 20x + 20^\circ = 180^\circ \]

### Step 6: Solve for \( x \)
Subtract \( 20^\circ \) from both sides:
\[ 20x = 160^\circ \]
Dividing both sides by \( 20 \):
\[ x = 8 \]

### Step 7: Find the Value of \( C \)
Now we can find \( C \) using \( x \):
\[ C = 3x + 20^\circ = 3(8) + 20^\circ = 24 + 20 = 44^\circ \]

### Conclusion
Thus, the third angle \( C \) is:
\[ \boxed{44^\circ} \]

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