Three angles of a triangle are \( (3 x+12)^{\circ},(2 x-38)^{\circ},(x+10)^{\circ} \). Find the value of each angle. In an isosceles triangle, the vertical angle in \( 40^{\circ} \). Find the value of base angles. Find the acute angles of an isosceles right angled triangle. In a triangle, one of the angles is greater than other by \( 20^{\circ} \) and the thir angle is \( 50^{\circ} \). Find the two angles.

To solve the problem about the three angles of the triangle, we start by setting up the equation based on the fact that the sum of angles in a triangle is \(180^\circ\): \[ (3x + 12) + (2x - 38) + (x + 10) = 180 \] Combining like terms gives: \[ 6x - 16 = 180 \] Solving for \(x\): \[ 6x = 196 \quad \Rightarrow \quad x = \frac{196}{6} \quad \Rightarrow \quad x = 32.67 \] Now, substituting \(x\) back into the angles: - Angle 1: \(3(32.67) + 12 = 108.01^\circ\) - Angle 2: \(2(32.67) - 38 = 27.34^\circ\) - Angle 3: \(32.67 + 10 = 42.67^\circ\) Rounding gives us the angles as approximately \(108^\circ\), \(27^\circ\), and \(43^\circ\). For the isosceles triangle with a vertical angle of \(40^\circ\), the base angles are equal. Therefore: \[ 40 + 2b = 180 \quad \Rightarrow \quad 2b = 140 \quad \Rightarrow \quad b = 70^\circ \] Thus, each base angle measures \(70^\circ\). In an isosceles right triangle, the two equal angles are each \(45^\circ\) since the right angle is \(90^\circ\) and the angles must add up to this together with the right angle. Lastly, in the triangle where one angle is \(50^\circ\) and one angle is \(20^\circ\) more than the other, we let the smaller angle be \(x\): \[ x + (x + 20) + 50 = 180 \] This simplifies down to: \[ 2x + 70 = 180 \quad \Rightarrow \quad 2x = 110 \quad \Rightarrow \quad x = 55 \] So, the angles are \(55^\circ\), \(75^\circ\) (which is \(55 + 20\)), and \(50^\circ\).