An isosceles triangle abc with the base bc is inscribed into a circle. find the measure of angles of it, if measure of arc bc=102°.

The measures of the angles of ΔABC are

m∠A = 51° , m∠B = 64.5 , m∠C = 64.5°

Step-by-step explanation:

* Lets revise some facts in the circle

- The measure of the circle is 360°

- Equal chords intercept equal arcs

- The measure of an inscribed angle equals half the measure

 of the intercepted arc

* Lets solve the problem

- ΔABC is isosceles withe base BC

∴ AB = AC

- The vertex angle is A

∵ ∠A is an inscribed angle subtended by arc BC

∴ m∠A = 1/2 the measure of arc BC

∵ The measure of arc BC = 102° ⇒ given

∴ m∠A = 1/2(102) = 51°

- In ΔABC

∵ AB = AC ⇒ proved

∴ m∠B = m∠C ⇒ base angles of isosceles Δ

- The sum of the interior angles in any Δ = 180°

∵ m∠A = 51 ⇒ proved

∴ m∠B = m∠C = (180 - 51) ÷ 2 = 129 ÷ 2 = 64.5°

* The measures of the angles of ΔABC are

  m∠A = 51° , m∠B = 64.5° , m∠C = 64.5°

m∠a is 90°

m∠b is 40°