71. An angle questions (synthetic/trigonometric proofs).
by Virgil Nicula, Jul 30, 2010, 2:56 AM
PP1. Let 
be an 
-right triangle with 
. Let 
, 
so that 
, 
. Ascertain 
.
PP2. Let be an -right triangle with . Let , so that , 
. Ascertain .
PP3. In-isoceles triangle the bisector line of 
meets 
at point 
. Show that 
.
Proof 1 (trigonometric). Denote
, where 
. Thus, 
and 
and 
.
Proof 2 (synthetic).Construct the equilateral
, where , 
are on the same side of 
. Rays 
meet circle 
at
. Observe that 
![$\frac{_1}{^2}\cdot \left[180^\circ - m(\angle YBX) \right]= 80^\circ$](//latex.artofproblemsolving.com/e/5/7/e57fe67268b1b75fa7b988a5718c8048d8536929.png)
and 
is 
-isosceles. The -isosceles 
, are congruent by symmetry
. On other hand, 
and 
is -isosceles 
.
Proof 3 (synthetic). Let
and 
such that 
and 
. Therefore, 
is a cyclic quadrilateral
and
, 
, i.e. the original isosceles triangle is now split up into four other isosceles triangles, two of which are congruent.
PP4. Given an
-isosceles triangle . Let 
and 
be two points such that 
. Suppose
that
, 
, 
, 
. Find the value of 
.
Proof 1 (trigonometric). Denote
and apply the well-known identity
![\[\sin\widehat{ODE}\cdot\sin\widehat{OAD}\cdot\sin\widehat{OBA}\cdot\sin\widehat{OEB}=\sin\widehat{OED}\cdot\sin\widehat{ODA}\cdot\sin\widehat{OAB}\cdot \sin\widehat{OBE}\]](//latex.artofproblemsolving.com/f/e/f/fefcb6bb33bdf00085be7adec974933bf2138d73.png)
![\[\sin (130^{\circ}-x)\cdot\sin 10^{\circ}\cdot\sin 60^{\circ}\cdot\sin 30^{\circ}=\sin x\cdot\sin 40^{\circ}\cdot\cos 20^{\circ}\cdot \sin 20^{\circ}\]](//latex.artofproblemsolving.com/7/4/2/7423573be864d351390592e3b309071bff843adf.png)
![\[\cos (40^{\circ}-x)\cdot\sin 10^{\circ}\sin 60^{\circ}=\sin x\cdot \sin^{2}40^{\circ}\]](//latex.artofproblemsolving.com/1/a/f/1af31eae7b5914dda21ce262f28234d3af78c70a.png)
![\[\cos (40^{\circ}-x)(\cos 50^{\circ}-\cos 70^{\circ})=\sin 40^{\circ}\left[\cos (40^{\circ}-x)-\cos (40^{\circ}+x)\right]\]](//latex.artofproblemsolving.com/f/2/f/f2f57665a46687f68fea66f80c2e6d30e24a848b.png)
![\[\sin 40^{\circ}\cos (40^{\circ}+x)=\cos (40^{\circ}-x)\sin 20^{\circ}\]](//latex.artofproblemsolving.com/f/5/e/f5e8de1c3be3b6bf6d882d875faabdbdd9ad042f.png)
![\[2\cos 20^{\circ}\cos (40^{\circ}+x)=\cos (40^{\circ}-x)\]](//latex.artofproblemsolving.com/e/2/e/e2e454880f9342699f63c0036cd7808fc5f080c4.png)
![\[\cos (60^{\circ}+x)+\cos (20^{\circ}+x)=\cos (40^{\circ}-x)\]](//latex.artofproblemsolving.com/3/e/d/3edafb318c45921bbb0ada33d79f7e9a808db72e.png)
![\[\cos (60^{\circ}+x)=\cos (40^{\circ}-x)-\cos (20^{\circ}+x)\]](//latex.artofproblemsolving.com/0/d/4/0d4009f557f762c40d1b86475a207c17fcc59193.png)
![\[\cos (60^{\circ}+x)=\sin (x-10^{\circ})\]](//latex.artofproblemsolving.com/b/b/3/bb3e6ab7a4d64ac7d4d50ae6427f58f51c89fa86.png)
![\[\sin (30^{\circ}-x)=\sin (x-10^{\circ})\]](//latex.artofproblemsolving.com/2/1/e/21ed3a037f38dbdd9c2d77283020eac69c9b9191.png)
![\[x=20^{\circ}\]](//latex.artofproblemsolving.com/f/3/4/f343f61b540d2dbaf656d76de6baa3c40277d46f.png)
Proof 2 (synthetic). Parallel to
through cuts at 
. 
cuts 
at 
. Bisector 
of 
cuts 
at 
. The triangle 
is F-isosceles and the
triangle
is Q-isosceles 
. 
is equilateral 
is F-isosceles with 
.
Proof 3 (synthetic). Take
so that 
and 
so that 
. See that 
is the circumcenter of 
and the circumcenter
of
we get 
. But we know that 
, so 
(to prove, take inside the triangle 
so that 
is equilateral,
see that
) . Next, see that 
is isosceles, so 
is isosceles. From the above sketch, 
and 
.
be an 
-right triangle with 
. Let 
, 
so that 
, 
. Ascertain 
.
. Apply in quadrilateral 
well-known relation :
.
PP2. Let
be an -right triangle with . Let , so that , 
. Ascertain .
. Apply same relation in quadrilateral 
PP3. In
-isoceles triangle the bisector line of 
meets 
at point 
. Show that 
.Proof 1 (trigonometric). Denote
, where 
. Thus, 
and 
and 
.Proof 2 (synthetic).Construct the equilateral
, where , 
are on the same side of 
. Rays 
meet circle 
at
. Observe that 
![$\frac{_1}{^2}\cdot \left[180^\circ - m(\angle YBX) \right]= 80^\circ$](http://latex.artofproblemsolving.com/e/5/7/e57fe67268b1b75fa7b988a5718c8048d8536929.png)
and 
is 
-isosceles. The -isosceles 
, are congruent by symmetry 
. On other hand, 
and 
is -isosceles 
.Proof 3 (synthetic). Let
and 
such that 
and 
. Therefore, 
is a cyclic quadrilateraland
, 
, i.e. the original isosceles triangle is now split up into four other isosceles triangles, two of which are congruent.PP4. Given an
-isosceles triangle . Let 
and 
be two points such that 
. Supposethat
, 
, 
, 
. Find the value of 
.Proof 1 (trigonometric). Denote
and apply the well-known identity![\[\sin\widehat{ODE}\cdot\sin\widehat{OAD}\cdot\sin\widehat{OBA}\cdot\sin\widehat{OEB}=\sin\widehat{OED}\cdot\sin\widehat{ODA}\cdot\sin\widehat{OAB}\cdot \sin\widehat{OBE}\]](http://latex.artofproblemsolving.com/f/e/f/fefcb6bb33bdf00085be7adec974933bf2138d73.png)
![\[\sin (130^{\circ}-x)\cdot\sin 10^{\circ}\cdot\sin 60^{\circ}\cdot\sin 30^{\circ}=\sin x\cdot\sin 40^{\circ}\cdot\cos 20^{\circ}\cdot \sin 20^{\circ}\]](http://latex.artofproblemsolving.com/7/4/2/7423573be864d351390592e3b309071bff843adf.png)
![\[\cos (40^{\circ}-x)\cdot\sin 10^{\circ}\sin 60^{\circ}=\sin x\cdot \sin^{2}40^{\circ}\]](http://latex.artofproblemsolving.com/1/a/f/1af31eae7b5914dda21ce262f28234d3af78c70a.png)
![\[\cos (40^{\circ}-x)(\cos 50^{\circ}-\cos 70^{\circ})=\sin 40^{\circ}\left[\cos (40^{\circ}-x)-\cos (40^{\circ}+x)\right]\]](http://latex.artofproblemsolving.com/f/2/f/f2f57665a46687f68fea66f80c2e6d30e24a848b.png)
![\[\sin 40^{\circ}\cos (40^{\circ}+x)=\cos (40^{\circ}-x)\sin 20^{\circ}\]](http://latex.artofproblemsolving.com/f/5/e/f5e8de1c3be3b6bf6d882d875faabdbdd9ad042f.png)
![\[2\cos 20^{\circ}\cos (40^{\circ}+x)=\cos (40^{\circ}-x)\]](http://latex.artofproblemsolving.com/e/2/e/e2e454880f9342699f63c0036cd7808fc5f080c4.png)
![\[\cos (60^{\circ}+x)+\cos (20^{\circ}+x)=\cos (40^{\circ}-x)\]](http://latex.artofproblemsolving.com/3/e/d/3edafb318c45921bbb0ada33d79f7e9a808db72e.png)
![\[\cos (60^{\circ}+x)=\cos (40^{\circ}-x)-\cos (20^{\circ}+x)\]](http://latex.artofproblemsolving.com/0/d/4/0d4009f557f762c40d1b86475a207c17fcc59193.png)
![\[\cos (60^{\circ}+x)=\sin (x-10^{\circ})\]](http://latex.artofproblemsolving.com/b/b/3/bb3e6ab7a4d64ac7d4d50ae6427f58f51c89fa86.png)
![\[\sin (30^{\circ}-x)=\sin (x-10^{\circ})\]](http://latex.artofproblemsolving.com/2/1/e/21ed3a037f38dbdd9c2d77283020eac69c9b9191.png)
![\[x=20^{\circ}\]](http://latex.artofproblemsolving.com/f/3/4/f343f61b540d2dbaf656d76de6baa3c40277d46f.png)
Proof 2 (synthetic). Parallel to
through cuts at 
. 
cuts 
at 
. Bisector 
of 
cuts 
at 
. The triangle 
is F-isosceles and thetriangle
is Q-isosceles 
. 
is equilateral 
is F-isosceles with 
.Proof 3 (synthetic). Take
so that 
and 
so that 
. See that 
is the circumcenter of 
and the circumcenterof
we get 
. But we know that 
, so 
(to prove, take inside the triangle 
so that 
is equilateral,see that
) . Next, see that 
is isosceles, so 
is isosceles. From the above sketch, 
and 
.This post has been edited 66 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:02 PM
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