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°
𝑇𝑆 = 𝑄𝑆 ⇒ 𝑥 = m (∠STQ) = m (∠SQT) =
= 75°
2
̅̅̅̅ and 𝑅𝑆
̅̅̅̅ are angle bisectors. If m (∠P) =
Example 3 In the figure, 𝑄𝑆
80°, how many degrees in ∠QSR?
Solution
m (∠QSR) = 180° − 𝑚(∠𝑆𝑄𝑅 ) − 𝑚(∠𝑆𝑅𝑄)
1
1
= 180° − 𝑚(∠𝑄 ) − 𝑚(∠𝑅)
= 180° −
= 180° −
2
1
2
1
2
1
2
[𝑚(∠𝑄) + 𝑚(∠𝑅 )]
[180° − 𝑚(∠𝑃)]
= 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 ∆RMQ is an isosceles
triangle. Hence, m (∠𝑀𝑅𝑄) = 32°. In triangle RMQ, ∠RMH is an
exterior angle. Thus,
m (∠RMH) = m (∠𝑀𝑅𝑄) + 𝑚 (∠𝑀𝑄𝑅 ) =32° + 32° = 64°
m (∠ HRM) = 90° – m (∠RMH) = 90° – 64° = 26°