Given the length of one side and the size of two angles.**Case 1:**

Suppose we wish to construct a triangle with one side 2 cm, and two angles 30° and 40°. Some possibilities are:

(ASA-Two angles and the included side)

These two triangles are congruent. i.e., \(\bigtriangleup{ABC} \cong \bigtriangleup{PRQ}\)

**Case 2:**

A corresponds to P, B corresponds to R and C corresponds to Q.

Again the two triangles are congruent. \(\bigtriangleup{ABC} \cong \bigtriangleup{PRQ}\)

(AAS-Two angles and the side opposite to one of them)

A corresponds to P, B corresponds to R, C corresponds to Q. Work out further possibilities for yourself. You will find that the triangles in each case are congruent.

## Angle-Side-Angle(ASA) Congruence condition:

If two triangles have **one side equal and two corresponding angles equal**, the two triangles are congruent.

### Angle-Side-Angle(ASA) Congruence condition Example 1:

Which of the following pairs of triangles are congruent?

1) \(\bigtriangleup{ABC}\): AB = 10 cm, \(\angle{A}\) = 40°, \(\angle{B}\) = 55°

\(\bigtriangleup{XYZ}\): XY = 10 cm, \(\angle{Y}\) = 40°, \(\angle{Z}\) = 85°

2) \(\bigtriangleup{PQR}\): PR = 5 cm, \(\angle{P}\) = 37°, \(\angle{R}\) = 64°

\(\bigtriangleup{DEF}\): DE = 5 cm, \(\angle{D}\) = 37°, \(\angle{F}\) = 64°**Solution:**

1) In \(\bigtriangleup{XYZ}\), we have

\(\angle{Y}\) = 40° and \(\angle{Z}\) = 85°

Therefore, \(\angle{X}\) = 180° – (\(\angle{Y} + \angle{Z}\)) = 180° – (40° + 85°) = 55°

Thus, in \(\bigtriangleup{ABC}\) and \(\bigtriangleup{XYZ}\), we have

AB = XY = 10 cm

\(\angle{A} = \angle{Y}\) = 40° and,

\(\angle{B} = \angle{X}\) = 55°

Therefore, by ASA congruence condition, we have \(\bigtriangleup{ABC} \cong \bigtriangleup{XYZ}\).

2) In \(\bigtriangleup{PQR}\) and \(\bigtriangleup{DEF}\), we have

PR = DE = 5 cm

\(\angle{P} = \angle{D}\) = 37° and,

\(\angle{R} = \angle{E}\) = 64°

Therefore, by ASA congruence condition, we have

\(\bigtriangleup{PQR} \cong \bigtriangleup{DFE}\) or \(\bigtriangleup{PRQ} \cong \bigtriangleup{DEF}\).