SAT / ACT GEOMETRY TRIANGLES The base angles of an isosceles triangle are equal Example 1 In a triangle RST, if RS = ST and m (∠T) = 70°, what is the value, in degrees, of angle S? Solution RS = ST ⇒ m (∠R) = m (∠T) = 70°; m (∠S) = 180° − (70° + 70°) = 40°. Example 2 In the figure, PQRS is a square and RST is an equilateral triangle. Find the value of x. Solution: m (∠TSQ) = 90° – m (∠RST) = 90° – 60° = 30° 180° − 30° ππ = ππ ⇒ π₯ = m (∠STQ) = m (∠SQT) = = 75° 2 Μ Μ Μ Μ and π π Μ Μ Μ Μ are angle bisectors. If m (∠P) = Example 3 In the figure, ππ 80°, how many degrees in ∠QSR? Solution m (∠QSR) = 180° − π(∠πππ ) − π(∠ππ π) 1 1 = 180° − π(∠π ) − π(∠π ) = 180° − = 180° − 2 1 2 1 2 1 2 [π(∠π) + π(∠π )] [180° − π(∠π)] = 180° − (180° − 80°) 2 = 180° − 50° = 130° Example 4 Solution Example 5 If the angles of a triangle are in the ratio of 2:3:5, what is the measure of the smallest angle? Solution Let 2x° be the smallest angle, then the other two angles would be 3x° and 5x° other two angles in degrees. 2x + 3x + 5x = 180° 10x = 180° x = 18° 2x = 36° Example 6 Solution Example 7 In a right triangle PQR, RH and RM are the altitude and the median to the hypotenuse. If m (∠Q) = 32°, find m (∠ HRM). Solution In a right triangle, the median to the hypotenuse equals halve the hypotenuse. Therefore, RM = PM = QM, and βRMQ is an isosceles triangle. Hence, m (∠ππ π) = 32°. In triangle RMQ, ∠RMH is an exterior angle. Thus, m (∠RMH) = m (∠ππ π) + π (∠πππ ) =32° + 32° = 64° m (∠ HRM) = 90° – m (∠RMH) = 90° – 64° = 26°