Let $ABC$ be an isosceles triangle such that $AB = AC$ and $\angle A = 20^o$. Let $M$ be the foot of the altitude from $C$ and let $N$ be a point on the side $AC$ such that $CN =\frac12 BC$. Determine the measure of the angle $AMN$.
Problem
Source: 2002 Romania JBMO TST2 p3
Tags: geometry, angles, Angle Chasing, equal segments
Ucchash
11.07.2020 15:29
Let,<AMN=x,<NMC=90-x and<MNC=20+x here x is an acute angle In BCM applying sine law CM=sin80/sin90 *BC.....(1) again in CMN applying sine law CM=sin(20+x)/sin(90-x) *CN...(2) from this two equation sin80/sin90=sin(20+x) /sin(90-x)*(CN/BC) or,sin80/sin30=sin(20+x)/sin(90-x) so,x=<AMN=60 Is this right?
Milana
02.08.2020 19:46
sunken rock wrote:
$60^\circ$
Use symmetry
Best regards, sunken rock How did you solve it?