d) If the angles of a triangle are in the ratio \( 2: 3: 5 \), find them. If two acute angles of a right angled triangle are in the ratio \( 1: 2 \), find them. If \( x^{\circ} \) is the exterior angle and \( 70^{\circ} \) and \( 50^{\circ} \) are its opposite interior angle of

To find the angles of a triangle when they are in a given ratio, we can use the fact that the sum of the angles in a triangle is always \( 180^{\circ} \). 1. For the first part, where the angles of a triangle are in the ratio \( 2: 3: 5 \): Let the angles be \( 2x, 3x, \) and \( 5x \) respectively. The sum of the angles in a triangle is \( 180^{\circ} \), so we have: \[ 2x + 3x + 5x = 180 \] \[ 10x = 180 \] \[ x = 18 \] Therefore, the angles of the triangle are \( 2x = 36^{\circ}, 3x = 54^{\circ}, \) and \( 5x = 90^{\circ} \). 2. For the second part, where two acute angles of a right-angled triangle are in the ratio \( 1: 2 \): Let the angles be \( x \) and \( 2x \) respectively. Since it is a right-angled triangle, the sum of the two acute angles is \( 90^{\circ} \), so we have: \[ x + 2x = 90 \] \[ 3x = 90 \] \[ x = 30 \] Therefore, the angles of the right-angled triangle are \( 30^{\circ} \) and \( 60^{\circ} \). 3. For the third part, where \( x^{\circ} \) is the exterior angle and \( 70^{\circ} \) and \( 50^{\circ} \) are its opposite interior angles: The sum of the interior and exterior angles of a triangle is always \( 180^{\circ} \). So, we have: \[ x + 70 + 50 = 180 \] \[ x + 120 = 180 \] \[ x = 60 \] Therefore, the exterior angle is \( 60^{\circ} \) and the opposite interior angles are \( 70^{\circ} \) and \( 50^{\circ} \).