A point $D$ is taken inside an isosceles triangle $ABC$ with base $AB$ and $\angle C=80^\circ$ such that $\angle DAB=10^\circ$ and $\angle DBA=20^\circ$. Compute $\angle ACD$.
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Tags: geometry
royliang
08.04.2021 03:00
Draw an accurate picture and then angle chase to find that $\angle ACD = 60^{\circ}$.
Krishijivi
21.07.2023 17:17
Let angle ACD=x, angle DCB=80°-x AC= BC, angle CAD=40°, angle BCD=30° IN ∆ DCB,angle DCB=70°+x BY SINE RULE, sin(70°+x)/BC=sin 30°/ CD BC/ CD=2 sin(70°+x) ∆ DCA,angle CDA=140°-x BY SINE RULE, sin(140°-x)/AC=sin 40°/ CD BC/ CD=sin(140°-x)/sin 40° 2sin( 70°+x)= sin(140°-x)/sin 40° So, if we solve this we shall get, cos(30°+x){2sin 10°-1}=0 sin 10°≠1/2, So cos(30°+x)=0= cos 90° ( x should be acute) 30°+x=90° x=60° So, angle ACD=60° # KRISHIJIVI
resources
21.07.2023 17:38
ive seen an exact same problem as this except they were asking for angle CDB instead I also used the solution above to solve it(trig bash w/ sum to product rules), but is there any elegant way? ive tried for so long with no avail
sunken rock
24.07.2023 13:54
Take $E\in AC|BE\bot AD$ and see that we are within the conditions of 'double angle lemma', so $\widehat{AED}=30^\circ\implies CDBE$ cyclic, thus $\widehat{ACD}=\widehat{DBE}=60^\circ$. Best regards, sunken rock
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