1) in the diagram below, we have , , , and . Find .
2) In the diagram below, we have . We also know that , , and . Find
3) In the diagram, angles and are right angles. If , , and , then what is ?
4) is a point on side of square such that and The extensions of and intersect at Find
5) In the diagram below, we have What is in degrees?
6) Segments and intersect at . We know that . Which angles below are equal to ?
1) Triangle PQR is similar to triangle PST , so
PQ / PS = PR / PT And PQ = 8 , PS = 8 + 6 , and PT = 10.5 .
8 / [8 + 6] = PR / 10.5 Multiply both sides of this equation by 10.5 .
10.5 * 8 / [8 + 6] = PR Simplify.
6 = PR
2) Notice that we have two parallel lines cut by a transversal, and their alternate angles are congruent. So we can be sure that triangle OMN is similar to triangle OQP , which means
OQ / OM = OP / ON And OQ = 6 , OM = 9 , and ON = 25 - OP
6 / 9 = OP / [25 - OP] Solve for OP .
10 = OP
3) Angle CAB is a part of both triangles, and both triangles have a right angle.
So triangle ABC is similar to triangle ADE by angle-angle similarity.
AB / AC = AD / AE And AB = 11 + 10 , AC = 35 , and AE = 11
[11 + 10] / 35 = AD / 11
11 * 21/35 = AD
4) Notice that angle DEF and angle CEB are vertical angles, angle BFD and angle FBC are alternate, and the other two are also alternate. So triangle DEF is similar to triangle CEB .
DF / DE = CB / CE Since CD and CB are sides of the same square, CD = CB = 2 + 5
DF / 2 = [2 + 5] / 5 I'll let you finish solving this for DF .
5) Triangle MNO is similar to triangle QPO , so ∠MNO = ∠QPO = 60°
There are 180° in every triangle. ∠RNO = ∠PQO = 180° - 60° - 66° = 54°
∠MNR = ∠MNO - ∠RNO = 60° - 54° = 6°