Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 46Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
x=30 degrees is also a solution to the problem
In Tr. DBC, DC/DB = sin(3x+30)/sin(30) = 2sin(3x+30) While Tr. ADC gives,DC/AD =sin(2x)/sin(x)Now AD=DB gives, 2sin(3x+30)=sin(2x)/sin(x) or2sin(3x+30)=2sin(x)cos(x)/sin(x) or sin(3x+30)=sin(90-x) or 3x+30=90-x or x=15Alternatively, sin(3x+30)=sin(90-x) may meansin(180 -(3x+30))=sin(90-x) giving us,180 -(3x+30) = 90- x giving us x=30Svvv is very correct in saying that 15 deg. and 30 deg. are both acceptable solutions to this problem.Ajit: firstname.lastname@example.org
Thanks for the comments.New condition: angle B is an obtuse angle.
Dear Antonio This problem is proving very difficult to crack thro Euclidean geometry alone. I tried to drop altitudes from B to DC and AC and work through the cyclic quadSecondly tried to find point M on DC such that D is the centre of ABM Tr. Thirdly found that if u find N on DC such that Tr. ANC is isoceles I found that < DBN is 30But I cannot clinch itAre any of my approaches right?
Dear Sumith,Some ideas: equilateral triangle, isosceles triangle, congruence ....
Let BE = h be an altitude of triangle ABC and let CD = dSo BD = DA = DE = CE = c/2CD is a median of triangle ABC, so d^2 = a^2/2 +b^2/2 -c^2/4 ....(1)Since triangles BDC and ADC are equal in area, easily1/2 da/2 = 1/2 bh/2 hence d = bh/a ....(2)Using Pythagoras h^2 = c^2 - (b-c/2)^2 = a^2 - (c/2)^2....(3)Eliminating d and h between these 3 equations,a^2/2 + b^2/2 - c^2/4 = (b^2/a^2)(a^2 - c^2/4).Simplifyinga^/2 - c^2/4 = b^2/2 - b^2c^2/(4a^2) which gives,2(a^2 - b^2) = (c^2/a^2)(a^2 -b^2)Since a NOT = b, a^2 = (c/2)cSo BC^2 = BD. BA and BC is therefore a tangent to circle ADC.Hence 2x = 30 and x = 15Sumith PeirisMoratuwaSri Lanka