In a `DeltaABC, /_A + /_B = 70^@` and `/_B + /_C = 130^@`, value of `/_A` is

Options

A. `20^@`
B. `50^@`
C. `110^@`
D. `30^@`

Answer

To find the value of angle A in triangle ABC, we can follow these steps:

### Step 1: Set up the equations based on the given information.
We know that:
1. \( \angle A + \angle B = 70^\circ \) (Equation 1)
2. \( \angle B + \angle C = 130^\circ \) (Equation 2)

### Step 2: Express angles in terms of one variable.
From Equation 1, we can express angle B in terms of angle A:
\[ \angle B = 70^\circ - \angle A \]

### Step 3: Substitute angle B into Equation 2.
Now, substitute \( \angle B \) from Equation 1 into Equation 2:
\[ (70^\circ - \angle A) + \angle C = 130^\circ \]
Simplifying this gives:
\[ \angle C = 130^\circ - (70^\circ - \angle A) \]
\[ \angle C = 130^\circ - 70^\circ + \angle A \]
\[ \angle C = 60^\circ + \angle A \]

### Step 4: Use the triangle angle sum property.
The sum of the angles in a triangle is always \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
Substituting the expressions for \( \angle B \) and \( \angle C \):
\[ \angle A + (70^\circ - \angle A) + (60^\circ + \angle A) = 180^\circ \]
This simplifies to:
\[ \angle A + 70^\circ - \angle A + 60^\circ + \angle A = 180^\circ \]
\[ 70^\circ + 60^\circ + \angle A = 180^\circ \]
\[ 130^\circ + \angle A = 180^\circ \]

### Step 5: Solve for angle A.
Subtract \( 130^\circ \) from both sides:
\[ \angle A = 180^\circ - 130^\circ \]
\[ \angle A = 50^\circ \]

Thus, the value of \( \angle A \) is \( 50^\circ \).

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