If QS−→ bisects ∠PQR, m∠PQS=7x−6, and m∠RQS=4x+15 . . . - Wyzant Set these two equal to each other (bisects means splits evenly into two parts, so that's why they are equal to each other) Solve for X, plug back into one of the equations to see how big 1 2 of the total angle is, then double that! 7x−6 = 4x+15 X = 7 plug this back into one of the equations to see one side is 43 degrees, double that for the answer, 86°
if QS bisects PQT, m SQT= (8x-25), m PQT= (9x+34), and m SQR . . . - Wyzant We know what ray QS bisects angle PQT If you draw out angle PQT and ray QS you can see that angle PQS is congruent to angle SQT Because of the definition of angle bisector This also means that angle SQT is half of PQT Because of this we know that we can set the measure of PQT equal to 2 times the measure of SQT
Triangle Proofs | Wyzant Ask An Expert You could prove triangle BAC congruent to triangle DAC by SAS CA bisects <BAD, AC perpendicular to BD Given m<BAC = m<DAC Definition of angle bisector <BCA = <DCA right angles Two perpendicular lines form right angles m<BCA = m<DCA = 90 Right angles are 90 degrees AC = AC Reflexive Property BAC congruent to DAC SAS Upvote • 0 Downvote Add
Given: AC bisects \angle BAD | Wyzant Ask An Expert Given: AC bisects \angle BADThis is what you need to do to construct the proof The bisection and the two segments that they want you to prove that they are equal make two triangles You need to first prove that these triangles, Triangle ABC and Triangle ADC are equal by Angle-Angle-side Then you can say that the two segments, BC and DC, are equal by "corresponding parts of congruent
Can you help with this question? | Wyzant Ask An Expert Ares J asked • 09 09 20 Can you help with this question? If ray QS bisects angle PQT, m angle SQT = (8×-25) degrees, m angle PQT = (9×+34) degrees, and m angle SQR = 112 degrees, find each: X = M angle PQS M angle PQT M angle TQR
f CDE is a straight angle, DE bisects GDH, m GDE = (8x 1) - Wyzant From the way the problem is worded: CDE is a straight angle made up of three angles: CDF of measure 43, FDG of unknown measure and GDE of measure 8x + 1 EDH is outside CDE and is of measure 6x + 15 Since DE bisects GDH to form GDE and EDH, those latter two angles are equal and thus 8x + 1 = 6x + 15 and x = 7 and substituting 7 for x, both angles are equal to 57 degrees Since CDF is 43 and
in triangle ABC, A is a right angle and D is a point on AC . . . - Wyzant The sum of the three angles in any triangle is 180 degrees Triangle ABC is a right triangle with the right angle at vertex A Point D is on the line AC so that the like BD bisects angle B So line BD divides the triangle ABC into two triangles, ABD and DBC, and the angle at vertex B is the same for both triangles Let's give that angle a name, X
GI−→ bisects ∠DGH so that m∠DGI is x−3 and m∠IGH . . . - Wyzant There are two parts to this problem First understand what a line that biscects an angle signifies GI--> bisects <DGH Now, understanding angles formed by a bisecting line are equivalent, solve for x in this single variable equation geometry problem <DGI = <IGH x-3 = 2x-13 2x-x = -3+13 x=10