dn 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 third angle is \( 50^{\circ} \). Find the two angles.

In an isosceles triangle, the two base angles are equal. Since the sum of the interior angles in a triangle is always \( 180^{\circ} \), you can find the base angles by subtracting the vertical angle from \( 180^{\circ} \) and then dividing by 2. So in this case: \((180^\circ - 40^\circ) / 2 = 70^\circ\). Thus, each base angle is \( 70^{\circ} \). For the isosceles right triangle, the right angle is \( 90^{\circ} \). Given that it is isosceles, the other two acute angles must be equal. So each of the two angles can be found by using the rule that the sum of angles in a triangle equals \( 180^{\circ} \): \((180^\circ - 90^\circ) / 2 = 45^\circ\). Therefore, each acute angle in an isosceles right triangle is \( 45^\circ \). To find the two angles where one is greater by \( 20^{\circ} \): Let the smaller angle be \( x \). Hence, the larger angle will be \( x + 20^\circ \). The third angle is given as \( 50^\circ \). Using the property that the sum of all three angles in a triangle is \( 180^\circ \), we can set up the equation: \( x + (x + 20^\circ) + 50^\circ = 180^\circ\). By simplifying this equation, you'll get \( 2x + 70^\circ = 180^\circ \). Solving it leads you to find \( 2x = 110^\circ \), and finally \( x = 55^\circ \). Thus, the two angles are \( 55^\circ \) and \( 75^\circ\) (the latter being \( 55^\circ + 20^\circ\)).