What Can You Conclude About Ac In Triangle Abc

For what difference of AB and AC is the a triangle ABC, given that BC is 3cm and angle C is 60 degrees? Is this possible?

The answer by Eeshaan Jain is correct, but it doesn’t always have to be right angled triangle, but to find the exact difference we must at least know one more side or angle.To solve this problem we have to use sine rule, which states [math]\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}[/math]wherea= BCb= CAc=AB are the sides of triangle and A,B,C are angles of triangle.From the given data we can conclude that[math]\frac{3}{sin A} = \frac{b}{sin B} = \frac{c}{\frac {\sqrt{3}}{2}}[/math]Solving the equations we get[math]c= \frac{3\sqrt{3}cosec A}{2}[/math][math]b= 3 cosec A sin B [/math]the difference would be [math]c[/math][math]-b = 3 cosec A[\frac{\sqrt 3}{2} - sin B][/math] if [math]B < 60 [/math]degrees or[math] A > 60 [/math]degreesand[math]b-c = 3 cosec A[sin B - \frac{\sqrt 3}{2}][/math] if [math]B > 60 [/math]degrees or [math]A < 60 [/math]degreesso A is 90 degrees then B is 30 degrees, soc-b = 3*1 ([math]\frac{\sqrt 3}{2} - \frac{1}{2}) =[/math] 1.09807 cmsor if B is 90 degrees then A is 30 degrees, so[math]b-c = [/math]3*2[math] [1 - \frac{\sqrt 3}{2}] = [/math] 0.80384 cmsand if both A and B are equal then the difference would be 0 because it becomes equilateral triangleAlso refer triangle inequality theorema+b>c, soa>c-bor a+c>b, soa>b-c Simplydifference of AB and AC is less than BC i.e., 3 cms

If I have the triangle ABC with AB=AC and the angle bisector of B is drawn so it intersects AC in D, how can I calculate the angle BAC? Such that BD+AD=BC.

Draw the point [math]E[/math] on the edge [math]BC[/math] so that [math]BE=BD[/math]. Then triangle [math]BDE[/math] is isosceles. Since [math]BD+AD = BC[/math] and [math]BE=BD[/math], we conclude that[math]CE = BC - BE = BC - BD = AD[/math]Furthermore, since [math]BD[/math] is the angle bisector of angle [math]\angle \, ABC [/math]and [math]AB = CA[/math],[math] [/math][math]\frac{CA}{CB} = \frac{AB}{CB} = \frac{AD}{CD} = \frac{CE}{CD}[/math]Hence, triangles [math]\Delta \, DEC[/math] and [math]\Delta\, ABC[/math] are similar, because they also share a common angle. However, triangle [math]\Delta \, ABC[/math] is isosceles, so triangle [math]\Delta \, DEC[/math] is also isosceles. Consequently,[math]\angle\, CDE = \angle \, DCE = \angle \, BCA = 2\,\alpha.[/math]Thus, angle[math]\angle \, DEC = 180 - (\angle \, CDE + \angle \, DCE) = 180 - 4\, \alpha.[/math]and so[math]\angle\, BED = 180 - \angle \, DEC = 4 \, \alpha[/math]However, triangle [math]\Delta \, BDE[/math] is isosceles, so[math]\angle \, BDE = \angle \, BED = 4\, \alpha[/math]Finally, angle[math]\angle \, DBE = \frac{1}{2}\, \angle \, ABC = \frac{1}{2}\, \angle \, BCA = \frac{1}{2}\, 2\, \alpha = \alpha.[/math]Since the angles in the triangle [math]\Delta\, BDE[/math] sum up to [math]180[/math][math]180 = \angle \, BED + \angle \, BDE + \angle \, DBE = 4\, \alpha + 4\, \alpha + \alpha = 9\, \alpha[/math]so[math]\angle \, DBC = \angle \, DBE = \alpha = \frac{180}{9} = 20[/math][math]\angle \, ABC = \angle \, BCA = 2\, \alpha = 40[/math]and thus[math]\angle \, BAC = 180 - (\angle \, ABC + \angle \, BCA) = 180 - 40 - 40 = 100[/math]

In a triangle ABC, M and N are on AC and BC respectively such that CM=CN. The intersecting point of AN and BM is O. If AO=BO and one of the angles of ABC is 40°, what are the measurements of the other angles of the triangle?

I believe triangle ABC must be isosceles. I suspect someone cleverer than I can come up with a simpler proof than mine; I’d like to see it. Meanwhile, see the (intentionally) misdrawn diagram below. (My many thanks to fellow Quoran John Cardan for explaining to me how I can get my own diagram into a Quora answer. A math diagram is worth 2^10 words.)Let’s suppose OM < ON (as in the diagram). Then, since AO = BO and m AM and b) m m CN which contradicts the given, CM = CN. A comparable argument works if OM > ON. Thus we must have OM = ON.Now if OM = ON, then triangles MBA and NAB are congruent by SAS from which we get c) m