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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 819.
Triangle ABC is isosceles, AC = 2ABcos40Using sine rule on triangle ACD, AC/sinx = CD/sin302ABcos40/sinx = CD/sin30cos40 = sinxx = 50
why x not= 130?I believe x=50 or 130.
In triangle ABC: AC = 2 AB cos(40) ---- (1)In triangle ACD:AC = CD sin(x)/sin(30) = 2 CD sin(x) ---- (2)AB = CD;Then by (1) and (2):sin(x) = cos(40) = sin(50)x = 50
Why x=50 only?I believe x=50 or 130.
To Anonymous: according to the figure angle x is acute, therefore x=50. If D' is on AD so that AB=BC=CD', then angle x is obtuse (x=130).
Antonio, did you judge the answer according to figure?
To Anonymous: I think the problem has two solutions. First, according to the figure, when the angle x is acute and the second, when the angle x is obtuse. Thanks.
Problem 819: Try to use elementary geometry (Euclid's Elements).
Let E be the circumcenter of ΔACD. Then∠CED = 2×∠CAD = 60°, CD = CE⇒ ΔCDE is equilateral⇒ AB = BC = CD = DE = CE = AE⇒ ABCE is a rhombus⇒ ∠AEC = ∠ABC = 100°⇒ ∠ADC = 1/2×∠AEC = 50°
draw the altitudes BM of triangle ABC,CH of triangle ACDCM=AC/2 (ABC is isoscele)CH=AC/2 (triangle 30-60-90)the right triangles BMC and CHD are congruentx=50
Costruind simetricul punctului C fata de dreapta AD si notandu-lcu P obtinem :ΔACP echilateral si ΔACP=ΔPCD(LLL);∠CDP = ∠ABC = 100°,AD mediatoarea seg AD =>AD bisectoarea ∠CDP => ∠CDA = 50° Prof Radu Ion,Sc.Gim.Bozioru,Buzau
Draw BE perpendicular to AC and CF perpendicular to AD. Join EF.BE bisects AC, since AB = BCE being the midpoint of the hypotenuse AC of the right triangle AFC, EA = EC = EF and consequently ΔEFC is equilateral.Now consider the right triangles BEC and DFC.By Pythagoras, DF^2 = CD^2 - CF^2 = BC^2 - CE^2 = BE*2.So DF = BE and consequently the right triangles CDF and CBE are congruent.Follows x = ∠FBC.Note that ΔABC is isosceles (with base angles FCB, FAB each 40°),Hence x = ∠FBC = 90° - ∠FCB = 90° - 40° = 50°
x=50deg if AD>AC; or x=130deg IF AD<AC.
Let BA=BC=CD=a and AC=dUsing sine rule,In triangle ABC, a/sin 40=d/sin 100…(1)In triangle ACD, a/sin 30=d/sin x…(2)Using (1),(2), we get, sin 100/sin40=sin x/sin 30 sin 80/sin40=sin x/sin 30 2(sin 40)(cos 40)/sin40=sin x/(1/2) cos 40=sin x sin (90-40)=sin x sin 50=sin x 50=x.
Let E be the point on AD such that AB= BE.ang. AEB= 700 (∆ ABE is isosceles) ang ABC= 1000= ang. ABE+ ang. EBC → ang EBC= 600Since BE=BC (by our construction) so, ang. BEC= ang. ECB = 1800-600/2= 600This implies ∆ BEC is equilateral. → CD= CE.so ang. X= ang. CED = 1800-(700+600) = 500□
Sorry for the extra zeros in the solution, they all refers degrees.Thank You.
En la figura adjunta resuelvo el problema sin usar CD...:Dhttp://www.subirimagenes.com/otros-ahorasi-8277920.htmlBy Tony García.
Draw altitudes BH and CG. Then by 30-60-90 Triangle and congruence x=50Sumith PeirisMoratuwaSri Lanka
Another proof. Find E on AD such that AB = BE. Then < EBC = 60 and Tr. EBC is equilateral, Tr. ECD is isoceles and x = 50. Sumith PeirisMoratuwaSri Lanka
Let D' on AD so that CD'=CD and <AD'C is obtuse and let B' be the circumcenter of tr. AD'C. Clearly tr CB'D' is equilateral and CB'=AB'=CD'=CB=AB, so B'=B, thus <AD'C=180-m(<ABC)/2=130 degs. and, as constructed, <ADC=<CD'D=180-m(<AD'C)=50 degs.Best regards