Properties of Triangles: Problems with Solutions

Solution:
30 + 40 < 80 so 30, 40, 80 can not be sides of a triangle.

Solution:
Each side must be less than the sum of the other two:
[tex]70+60=130\gt 100, 70+100=170\gt 60, 60+100=160\gt 70[/tex]
Answer: yes.

Solution:
The sum of the angles of a triangle is [tex]180^\circ[/tex].
[tex]30^\circ+40^\circ+100^\circ=170^\circ\ne 180^\circ[/tex]
Answer: no.

Solution:
The sum of the angles of a triangle is [tex]180^\circ[/tex].
[tex]20^\circ+60^\circ+95^\circ=175^\circ\ne 180^\circ[/tex]
Answer: no.

Solution:
The sum of the angles of a triangle is [tex]180^\circ[/tex].
[tex]50^\circ+53^\circ+77^\circ=180^\circ[/tex]
Answer: yes.

Solution:
The sum of the angles of a triangle is 180°.
Let the unknown angle be 'x' => 31 + 60 + x = 180
=> 91 + x = 180
=> x = 180 - 91
=> x = 89
The third angle is 89°

Solution:
The interior angle next to the [tex]82^\circ[/tex] exterior angle is [tex]180^\circ-82^\circ=98^\circ[/tex]. It is obtuse, and a triangle has at most one obtuse angle, so it cannot be one of the two equal base angles. Hence the vertex angle is [tex]98^\circ[/tex] and each base angle is:
[tex]\dfrac{180^\circ-98^\circ}{2}=41^\circ[/tex]
[tex]98^\circ-41^\circ=57^\circ[/tex]

Solution:
The exterior angles (one at each vertex) add up to [tex]360^\circ[/tex]:
[tex]4k+3k+2k=360^\circ \Rightarrow k=40^\circ[/tex]
The exterior angles are [tex]160^\circ[/tex], [tex]120^\circ[/tex] and [tex]80^\circ[/tex], so the interior angles are:
[tex]180^\circ-160^\circ=20^\circ, 180^\circ-120^\circ=60^\circ, 180^\circ-80^\circ=100^\circ[/tex]

Solution:
[tex]\angle B+(\angle B+40^\circ)+(\angle B+20^\circ)=180^\circ \Rightarrow \angle B=40^\circ, \angle A=80^\circ, \angle C=60^\circ[/tex]
In a triangle the larger side lies opposite the larger angle. Since [tex]\angle A\gt \angle C\gt \angle B[/tex]:
[tex]BC\gt AB\gt AC[/tex]

Solution:
If the base is [tex]x[/tex] and the legs are [tex]x+6[/tex]:
[tex]x+2(x+6)=24 \Rightarrow x=4[/tex]
The sides are [tex]10, 10, 4[/tex], and [tex]4+10\gt 10[/tex], so this triangle exists.
If the legs are [tex]y[/tex] and the base is [tex]y+6[/tex]:
[tex]2y+y+6=24 \Rightarrow y=6[/tex]
The sides would be [tex]6, 6, 12[/tex], but [tex]6+6=12[/tex], so such a triangle does not exist.

Solution:
Let the leg be [tex]b[/tex]; then the base is [tex]b+5[/tex]. By the triangle inequality:
[tex]b+b\gt b+5 \Rightarrow b\gt 5 \Rightarrow b\ge 6[/tex]
The smallest base is [tex]6+5=11[/tex] (triangle [tex]6, 6, 11[/tex]).

Solution:
In the right triangle [tex]AHC[/tex] the leg [tex]CH[/tex] lies opposite the [tex]30^\circ[/tex] angle, so it is half of the hypotenuse [tex]AC[/tex]:
[tex]AC=2CH=8[/tex]
[tex]P=AB+BC+AC=20+8=28[/tex]

Solution:
A point at equal distances from the arms of an angle lies on its bisector, so [tex]CN[/tex] bisects [tex]\angle ACB[/tex]. Let [tex]P[/tex] be the foot of the perpendicular from [tex]N[/tex] to [tex]CA[/tex]. In the right triangle [tex]CPN[/tex] the leg [tex]NP=13[/tex] is half of the hypotenuse [tex]CN=26[/tex], so:
[tex]\angle NCP=30^\circ \Rightarrow \angle ACB=2\times 30^\circ=60^\circ[/tex]

Solution:
Triangle [tex]AMC[/tex] is isosceles, and its exterior angle at [tex]A[/tex] is [tex]\angle BAC=60^\circ[/tex]. It equals the sum of the two equal base angles, so [tex]\angle ACM=30^\circ[/tex]. In the same way [tex]\angle BCN=\frac{50^\circ}{2}=25^\circ[/tex]. Also [tex]\angle ACB=180^\circ-60^\circ-50^\circ=70^\circ[/tex]:
[tex]\angle MCN=30^\circ+70^\circ+25^\circ=125^\circ[/tex]
In general, [tex]\angle MCN=90^\circ+\frac{\angle C}{2}[/tex].

Solution:
[tex]\angle BAC+2\angle BAC=90^\circ \Rightarrow \angle BAC=30^\circ, \angle ACB=60^\circ[/tex]
The leg [tex]BC[/tex] lies opposite the [tex]30^\circ[/tex] angle, so [tex]AC=2BC=20[/tex]. The median to the hypotenuse is half of the hypotenuse:
[tex]BM=\dfrac{AC}{2}=10[/tex]

Solution:
[tex]CB=CA=CM[/tex], so triangles [tex]ABC[/tex] and [tex]CBM[/tex] are isosceles. Let [tex]\angle CAB=\angle CBA=x[/tex] and [tex]\angle CBM=\angle CMB=y[/tex]. Since [tex]C[/tex] lies on segment [tex]AM[/tex], [tex]\angle ABM=x+y[/tex]. In triangle [tex]ABM[/tex]:
[tex]x+(x+y)+y=180^\circ \Rightarrow x+y=90^\circ \Rightarrow \angle ABM=90^\circ[/tex]