Solving triangle ABC?
Because the triangle is given with "SAS" information, you use the Law of Cosines: AB² = 11² + 23² - 2·11·23·cos 120° AB² = 650 - 506·- ½ AB² = 650 + 253 AB² = 903 AB = √903 = 30.050 To find m∡A, you can use Law of Sines: sin A/ 23 = sin 120°/30.05 sin A = 23·√3/2 / 30.05 sin A = .6628 A = arcsin(.6628) = 41.5° m∡B = 180° - 120° - 41.5° = 18.5°
How do I solve 8AE^2 = 3AC^2+5AD^2 in similar triangles?
The problem is incomplete in that what are D and E is not specified.Kindly avoid posting incomplete problem/s.I have seen this problem elsewhere, where it has been mentioned that D and E are points of trisection of the side BC and angle ABC is 90°.So if we take BC = 3x, then BD = DE = EC = x. Applying Pythagoras theorem to each of the right angled triangles ABD, ABE and ABC successively we have AD^2 = AB^2 + BD^2 => AD^2 = AB^2 + x^2 ……………….(1) Similarly, AE^2 = AB^2 +4x^2…..…………..(2) and also we have AC^2 = AB^2 + 9x^2 ……...(3).Now from (2), 8 AE^2 = 8 AB^2 + 32 x^2.From (1) and (3), 3AC^2 + 5 AD^2 = (3AB^2 + 27x^2) + (5AB^2+5x^2) = 8 AB^2 + 32 x^2.Hence, 8 AE^2 = 3 AC^2 + 5 AD^2.
In triangle ABC, angle A=90°, angle B=30°, BC=8cm. Find the length of side AB and AC?
angle A = 90° and angle B 30°=> angle C = 90° - 30° = 60°=> BC is the hyp = 8 cm longAC = 8/2 =4cm (sin30° = AC/hp => 8sin30°=ACAC = 8×1/2 = 4AB = √(BC^2 - AC^2) = √(8^2 - 4^2) = √(64-16) = 6.93cmAC = 4 cmAB = 6.93 cmHappy new year
There is a triangle ABC, with two points D and E on BC and BD=EC. Can you prove AB+AC>AD+AE?
Here is the answer.Draw [math]DF\parallel AC[/math] and [math]BF\parallel AE[/math]. The two lines intersect at point F. Line [math]AB[/math] and [math]DF[/math] intersect at point G.So we can prove that [math]DF=AC, BF=AE[/math]. Therefore [math]AG+DG>AD, AG+DG-AD>0[/math], [math]BG+FG>FB, BG+FG-FB>0[/math]Add the two equations, we have [math]AG+DG-AD+BG+FG-FB>0[/math], so [math]AB+FD>AD+BF[/math], [math]AB+AC>AD+AE[/math].Oh, here's another way:Draw [math]AF\parallel CB[/math] and [math]EF\parallel CA[/math]. They intersect at point F. Connect points B and F. As we can see, quadrilateral FECA is a parallelogram and [math]EF=AC[/math], [math]AF=CE[/math].As a result, [math]BD=AF[/math] and we also know [math]BD\parallel AF[/math], so quadrilateral FBDA is also a pqrqllelogram like FECA. Therefore [math]FB=AD[/math].In [math]\triangle AGE[/math], [math]AG+EG>AE[/math].In [math]\triangle BFG[/math], [math]BG+FG>FB[/math].So [math]AG+EG+BG+FG>AE+FB[/math], therefore [math]AB+AC>AD+AE[/math].It's just translated from the exam in Beijing.
In Triangle ABC, AD bisects Angle BAC. If AC=10cm, AB=14cm and BC=16cm, find BD?
1) You have to call a theorem and a property here for solving this in a simple way: i) Theorem: In a triangle, the bisector of an angle divides the opposite base in ratio equal to that of the other two sides containing the angle. Here, in triangle ABC, AD bisects angle BAC; So by the above theorem, AC/AB = CD/BD. ii) Property: If real numbers, a,b,c & d are such that, a/b = c/d, then (a+b)/b = (c+d)/d [This property is called 'compenendo' property of proportions] [Proof of this is very simple: Just add 1 to both sides of a/b = c/d; then taking LCM and simplifying we get (a+b)/b = (c+d)/d] iii) So from (i) & (ii) of above, (DC + BD)/BD = (AC+AB)/AB ==> BC/BD = (AC + AB)/AB ==> BD = (BC x AB)/(AC + AB) Substituitng the given values, BD = (16 x 14)/24 = 28/3 = 9 and (1/3) cm
Solve D = ABC for B.?
B = D/AC
ABC is an isosceles triangle with base BC inscribed in circle O of radius 5 cm. If BC = 8 cm, find AB?
OA = OB = OC = 5 BC = 8 Drop perpendicular from O to BC at point D Triangle ODB is right-angled at D, with OB = 5 and DB = 8/2 = 4 OD² + DB² = OB² OD² + 4² = 5² OD² = 25 - 16 = 9 OD = 3 Now triangle ADB is right angled at D, with AD = OA + OD = 5 + 3 = 8 AB² = AD² + DB² AB² = 8² + 4² AB² = 64 + 16 AB² = 80 AB = 4√5 Mαthmφm
In triangle ABC, angle ABC = 45 degrees. Point D is on segment BC such that 2|BD| = |CD| and angle DAB= 15 degrees. What is the angle ACB?
This is a classic constructions problem.Construct point [math]E[/math] such that [math]\angle EBC = \angle ABC, \angle EAD = \angle BAD[/math], and [math]E[/math] is on the opposite side of [math]BC[/math] to [math]A[/math]. Thus, we have [math]D[/math] is the incentre of [math]ABE[/math], which we note is a [math]30-60-90[/math] triangle, so [math]\frac{AE}{EB} = 2[/math].Extend [math]ED[/math] to intersect [math]AB[/math] at [math]F[/math]. By angle bisector theorem, [math]\frac{AF}{AB} = \frac{AE}{EB} = 2[/math]. Furthermore, [math]\frac{CD}{DB}=2[/math], so by [math]SAS[/math], [math]BFD \sim BAC[/math]. However, [math]\angle BDF = 75^{\circ}[/math] by a simple angle chase, so [math]\angle ACB = 75^{\circ}[/math].
In an equilateral triangle ABC, the side BC is trisected at D then 9AD square is?
GIVEN: Equilateral triangle ABC , SO each side is ‘a' . BC is trisected , so BD = DE = EC = a/3 & AM is perpendicular to BC => M is mid point of BC.=> DM = DE/2 = (a/3)/2 = a/6If < ADB is obtuse, then < ADC will be acuteBy Extension of Pythagoras theoremIn obtuse tri ADBAB² = AD² + BD² + 2BD*DM ……….(1)In acute triangle ADCAC² = AD² + DC² - 2DC*DM ………(2)By adding (1) & (2)AB² + AC² = 2AD² + BD² + DC² + 2BD*DM - 2DC*DMNow, write every segment in the form of ‘a'2a² = 2AD² + (a/3)² + (2a/3)² + 2* a/3 * a/6 - 2*2a/3*a/6=> 2a² = 2AD² + a²/9 + 4a²/9 + a²/9 - 2a²/9=> 2a² = 2AD² +4a²/9=> 2AD² = 2a² - 4a²/9=> 2AD² = 14a²/9=> 18AD² = 14a²=> 9AD² = 7a²=> 9AD² = 7 side²