Law of Cosines: Problems with Solutions

Solution:
According to the Law of Cosines $b^{2}=a^{2}+c^{2}-2ac\cos \beta $
then $b^{2}=12^{2}+10^{2}-2\cdot 12 \cdot 10\cos 26\frac{\pi }{180}=144+100-240(0.90)=28$

So $b=\sqrt{28}=2\sqrt{7}$

Solution:
By the Law of Cosines: [tex]BC^2=AC^2+AB^2-2 \times AC \times AB \times \cos\alpha[/tex].
Hence [tex]\cos\alpha=\frac{AC^2+AB^2-BC^2}{2 \times AC \times AB}=\frac{36+81-49}{2 \times 6 \times 9}=\frac{68}{108}=\frac{17}{27}\approx 0.6296[/tex].
Therefore [tex]\alpha=\arccos\frac{17}{27}\approx 50.98^\circ[/tex].

Solution:
By the law of cosines for triangle [tex]ABC[/tex], we have [tex]AB^2=AC^2+BC^2-2AC\cdot BC \cdot \cos(\angle ACB)[/tex]. Rearranging the terms, we get [tex]2AC \cdot BC \cdot \cos(\angle ACB)=AC^2+BC^2-AB^2[/tex]. Dividing both sides by [tex]2AC \cdot BC[/tex] gets us [tex]\cos(\angle ACB)=\frac{AC^2+BC^2-AB^2}{2AC \cdot BC}=\frac{3^2+5^2-6^2}{2 \cdot 3 \cdot 5}=\frac{9+25-36}{30}=\frac{-2}{30}=-\frac{1}{15}[/tex]

Solution:
By the law of cosines, we have [tex]AB^2=AC^2+BC^2-2.AC.BC.\cos\angle ACB[/tex]. Substituting [tex]AC[/tex], [tex]BC[/tex] and the angle with their values, we get: [tex]AB^2=12^2+10^2-2\cdot 12\cdot 10\cdot \cos(60^{\circ})[/tex], or [tex]AB^2=144+100-2\cdot 120 . \frac{1}{2}[/tex]. By performing the arithmetical operations on the right side, we get [tex]AB^2=124[/tex].

Solution:
The largest angle lies opposite the longest side, [tex]19[/tex]. Call it [tex]\gamma[/tex].
By the Law of Cosines: [tex]19^2=5^2+16^2-2 \times 5 \times 16 \times \cos\gamma[/tex].
[tex]361=281-160\cos\gamma \Rightarrow \cos\gamma=-\frac{1}{2}[/tex].
Therefore [tex]\gamma=120^\circ[/tex].

Solution:
In a parallelogram [tex]BC=AD=7[/tex] and [tex]\angle ABC=180^\circ-60^\circ=120^\circ[/tex].
By the Law of Cosines in [tex]\triangle ABC[/tex]:
[tex]AC^2=AB^2+BC^2-2 \times AB \times BC \times \cos 120^\circ=64+49-2 \times 8 \times 7 \times \left(-\frac{1}{2}\right)=113+56=169[/tex].
Therefore [tex]AC=13[/tex].

Solution:
By the Law of Cosines in [tex]\triangle ABC[/tex]:
[tex]\cos\angle BAC=\frac{AB^2+AC^2-BC^2}{2 \times AB \times AC}=\frac{81+36-144}{2 \times 9 \times 6}=\frac{-27}{108}=-\frac{1}{4}[/tex].
By the Law of Cosines in [tex]\triangle ADC[/tex]:
[tex]CD^2=AC^2+AD^2-2 \times AC \times AD \times \cos\angle BAC=36+16-2 \times 6 \times 4 \times \left(-\frac{1}{4}\right)=52+12=64[/tex].
Therefore [tex]CD=8[/tex].

Solution:
By the Law of Cosines in [tex]\triangle ABC[/tex]:
[tex]BC^2=AB^2+AC^2-2 \times AB \times AC \times \cos 45^\circ=32+1-2 \times 4\sqrt{2} \times 1 \times \frac{\sqrt{2}}{2}=33-8=25[/tex].
Hence [tex]BC=5[/tex].
The perimeter is [tex]P=4\sqrt{2}+1+5=6+4\sqrt{2}[/tex].

Solution:
The largest angle [tex]\gamma[/tex] is opposite the longest side, [tex]12[/tex].
By the Law of Cosines: [tex]\cos\gamma=\frac{7^2+8^2-12^2}{2 \times 7 \times 8}=\frac{49+64-144}{112}=-\frac{31}{112}[/tex].
Since [tex]\cos\gamma<0[/tex], the angle [tex]\gamma[/tex] is obtuse, so the triangle is obtuse.

Solution:
For [tex]c=9[/tex], [tex]c=10[/tex] and [tex]c=13[/tex] the longest side is [tex]c[/tex], and [tex]\cos\gamma=\frac{36+64-c^2}{2 \times 6 \times 8}=\frac{100-c^2}{96}[/tex].
For [tex]c=9[/tex]: [tex]\cos\gamma>0[/tex], so the triangle is acute.
For [tex]c=10[/tex]: [tex]\cos\gamma=0[/tex], so the triangle is right.
For [tex]c=13[/tex]: [tex]\cos\gamma=-\frac{69}{96}<0[/tex] and [tex]6+8>13[/tex], so the triangle exists and is obtuse.
For [tex]c=14[/tex]: [tex]6+8=14[/tex], so no such triangle exists.
Answer: [tex]c=13[/tex].

Solution:
Let [tex]AC=x[/tex]. By the Law of Cosines for the angle [tex]BAC[/tex]:
[tex]BC^2=AB^2+AC^2-2 \times AB \times AC \times \cos 60^\circ \Rightarrow 49=25+x^2-5x[/tex].
[tex]x^2-5x-24=0 \Rightarrow x_1=8;\ x_2=-3[/tex] (rejected, since a length is positive).
Therefore [tex]AC=8[/tex].

Solution:
Let [tex]\alpha[/tex] be the acute angle of the parallelogram. The diagonals lie opposite the angles [tex]\alpha[/tex] and [tex]180^\circ-\alpha[/tex]:
[tex]d_1^2=a^2+b^2-2ab\cos\alpha[/tex], [tex]d_2^2=a^2+b^2+2ab\cos\alpha[/tex].
Adding the two equations: [tex]d_1^2+d_2^2=2(a^2+b^2)[/tex].
[tex]18^2+d^2=2(14^2+8^2) \Rightarrow 324+d^2=520 \Rightarrow d^2=196[/tex].
The other diagonal is [tex]14[/tex].

Solution:
Let [tex]M[/tex] be the midpoint of [tex]AB[/tex]. Then [tex]AM=6[/tex].
By the Law of Cosines in [tex]\triangle AMC[/tex]:
[tex]CM^2=AC^2+AM^2-2 \times AC \times AM \times \cos 60^\circ=49+36-2 \times 7 \times 6 \times \frac{1}{2}=43[/tex].
Hence [tex]CM=\sqrt{43}[/tex].

Solution:
Let [tex]AB=10[/tex] be the base, [tex]AC=BC=b[/tex], and let [tex]M[/tex] be the midpoint of [tex]BC[/tex].
The altitude to the base bisects it, so [tex]\cos\angle ABC=\frac{5}{b}[/tex].
By the Law of Cosines in [tex]\triangle ABM[/tex]:
[tex]AM^2=AB^2+BM^2-2 \times AB \times BM \times \cos\angle ABC=100+\frac{b^2}{4}-2 \times 10 \times \frac{b}{2} \times \frac{5}{b}=50+\frac{b^2}{4}[/tex].
[tex]66=50+\frac{b^2}{4} \Rightarrow b^2=64[/tex].
Each leg is [tex]8[/tex].

Solution:
By the Law of Cosines, the third side satisfies [tex]c^2=25+64-2 \times 5 \times 8 \times \frac{1}{2}=49[/tex], so [tex]c=7[/tex].
Side [tex]c[/tex] is opposite the [tex]60^\circ[/tex] angle. By the Law of Sines:
[tex]R=\frac{c}{2\sin 60^\circ}=\frac{7}{\sqrt{3}}=\frac{7\sqrt{3}}{3}[/tex].

Solution:
By the Law of Cosines: [tex]\cos\angle BAC=\frac{AB^2+AC^2-BC^2}{2 \times AB \times AC}=\frac{196+169-225}{2 \times 14 \times 13}=\frac{140}{364}=\frac{5}{13}[/tex].
[tex]\sin\angle BAC=\sqrt{1-\frac{25}{169}}=\frac{12}{13}[/tex] (the sine of an angle of a triangle is positive).

Solution:
[tex]\text{Area}=\frac{1}{2} \times AC \times BC \times \sin 60^\circ \Rightarrow 6\sqrt{3}=\frac{1}{2} \times 3 \times BC \times \frac{\sqrt{3}}{2} \Rightarrow BC=8[/tex].
By the Law of Cosines: [tex]AB^2=9+64-2 \times 3 \times 8 \times \frac{1}{2}=49[/tex], so [tex]AB=7[/tex].

Solution:
The diagonals bisect each other, so the side [tex]10[/tex] and the half-diagonals [tex]10[/tex] and [tex]12[/tex] form a triangle.
Let [tex]\varphi[/tex] be the angle between the half-diagonals, opposite the side [tex]10[/tex]:
[tex]10^2=10^2+12^2-2 \times 10 \times 12 \times \cos\varphi \Rightarrow 240\cos\varphi=144 \Rightarrow \cos\varphi=\frac{3}{5}[/tex].
Since [tex]\cos\varphi>0[/tex], [tex]\varphi[/tex] is the acute angle between the diagonals. Answer: [tex]\frac{3}{5}[/tex].

Solution:
The half-diagonals are [tex]2\sqrt{3}[/tex] and [tex]4[/tex], and the angles between them are [tex]150^\circ[/tex] and [tex]30^\circ[/tex].
[tex]a^2=12+16-2 \times 2\sqrt{3} \times 4 \times \cos 150^\circ=28+24=52[/tex], [tex]a=2\sqrt{13}[/tex].
[tex]b^2=12+16-2 \times 2\sqrt{3} \times 4 \times \cos 30^\circ=28-24=4[/tex], [tex]b=2[/tex].

Solution:
Let the other two sides be [tex]x[/tex] and [tex]x+7[/tex]. The side [tex]13[/tex] is opposite the [tex]60^\circ[/tex] angle:
[tex]13^2=x^2+(x+7)^2-2x(x+7)\cos 60^\circ=x^2+7x+49[/tex].
[tex]x^2+7x-120=0 \Rightarrow x_1=8;\ x_2=-15[/tex] (rejected).
The sides are [tex]8[/tex] and [tex]15[/tex].

Solution:
Let the sides be [tex]a[/tex] and [tex]2a[/tex]. By the Law of Cosines, the third side is [tex]c^2=a^2+4a^2-2 \times a \times 2a \times \frac{1}{2}=3a^2[/tex], [tex]c=a\sqrt{3}[/tex].
Then [tex]a^2+c^2=a^2+3a^2=4a^2=(2a)^2[/tex], so the angle opposite the side [tex]2a[/tex] is a right angle.
The triangle is a right triangle.

Solution:
Let the triangle be [tex]ABC[/tex], [tex]M \in AB[/tex], [tex]AM=1[/tex], and let [tex]O[/tex] be its center.
[tex]OA=R=\frac{6}{\sqrt{3}}=2\sqrt{3}[/tex], and [tex]AO[/tex] is an angle bisector, so [tex]\angle OAM=30^\circ[/tex].
By the Law of Cosines in [tex]\triangle AOM[/tex]:
[tex]OM^2=12+1-2 \times 2\sqrt{3} \times 1 \times \frac{\sqrt{3}}{2}=13-6=7[/tex].
Therefore [tex]OM=\sqrt{7}[/tex].

Solution:
[tex]\angle ABC=180^\circ-\angle BAD[/tex], so [tex]\cos\angle ABC=-\frac{7}{25}[/tex].
From [tex]\triangle ABC[/tex]: [tex]AC^2=a^2+a^2+2a^2 \times \frac{7}{25}=\frac{64a^2}{25} \Rightarrow 64=\frac{64a^2}{25} \Rightarrow a=5[/tex].
From [tex]\triangle ABD[/tex]: [tex]BD^2=2a^2-2a^2 \times \frac{7}{25}=50-14=36[/tex], [tex]BD=6[/tex].

Solution:
By the Law of Cosines: [tex]AC^2=25+64-2 \times 5 \times 8 \times \frac{1}{2}=49[/tex], [tex]AC=7[/tex].
The area is [tex]S=\frac{1}{2} \times 5 \times 8 \times \sin 60^\circ=10\sqrt{3}[/tex], and the semiperimeter is [tex]s=\frac{5+8+7}{2}=10[/tex].
[tex]r=\frac{S}{s}=\sqrt{3}[/tex].

Solution:
By the Law of Cosines for the angle [tex]\alpha[/tex]: [tex]a^2=b^2+c^2-2bc\cos\alpha[/tex].
[tex]16=25+c^2-5c \Rightarrow c^2-5c+9=0[/tex].
The discriminant is [tex]D=25-36=-11<0[/tex], so the equation has no real roots.
No such triangle exists. (Indeed, the distance from [tex]C[/tex] to the line [tex]AB[/tex] is [tex]b\sin\alpha=\frac{5\sqrt{3}}{2}>4=a[/tex].)

Solution:
The altitude to the base is also a median, so each leg is [tex]b=\sqrt{2^2+4^2}=2\sqrt{5}[/tex].
By the Law of Cosines for the vertex angle [tex]\gamma[/tex]:
[tex]\cos\gamma=\frac{b^2+b^2-4^2}{2b^2}=\frac{20+20-16}{40}=\frac{3}{5}[/tex].

Solution:
[tex]\gamma=180^\circ-(\alpha+\beta)[/tex], so [tex]\cos\gamma=-\cos(\alpha+\beta)=-\frac{1}{3}[/tex].
By the Law of Cosines: [tex]AB^2=9+16-2 \times 3 \times 4 \times \left(-\frac{1}{3}\right)=25+8=33[/tex].
Therefore [tex]AB=\sqrt{33}[/tex].

Solution:
By the Law of Cosines: [tex]AB^2=9+36-2 \times 3 \times 6 \times \left(-\frac{1}{2}\right)=63[/tex], [tex]AB=3\sqrt{7}[/tex].
The area is [tex]S=\frac{1}{2} \times 3 \times 6 \times \sin 120^\circ=\frac{9\sqrt{3}}{2}[/tex].
[tex]CH=\frac{2S}{AB}=\frac{9\sqrt{3}}{3\sqrt{7}}=\frac{3\sqrt{21}}{7}[/tex].

Solution:
The angle [tex]\beta[/tex] is obtuse, so [tex]\cos\beta=-\sqrt{1-\frac{16}{25}}=-\frac{3}{5}[/tex].
By the Law of Cosines: [tex]b^2=a^2+c^2-2ac\cos\beta=225+49+2 \times 15 \times 7 \times \frac{3}{5}=274+126=400[/tex].
Therefore [tex]b=20[/tex].

Solution:
The sides are positive, so [tex]0<\sin x<1[/tex] and [tex]0<\cos x<1[/tex], i.e. the longest side is [tex]1[/tex].
For the angle [tex]\varphi[/tex] opposite it: [tex]\cos\varphi=\frac{\sin^2 x+\cos^2 x-1}{2\sin x\cos x}=0[/tex].
Therefore [tex]\varphi=90^\circ[/tex].

Solution:
Applying the Law of Cosines to two adjacent angles of the parallelogram gives [tex]d_1^2+d_2^2=2(a^2+b^2)[/tex].
Let [tex]d_1=6k[/tex], [tex]d_2=7k[/tex]: [tex]36k^2+49k^2=2(121+49) \Rightarrow 85k^2=340 \Rightarrow k=2[/tex].
The diagonals are [tex]12[/tex] and [tex]14[/tex].

Solution:
By the Law of Cosines in [tex]\triangle ABD[/tex] [tex](AB=8,\ AD=5,\ BD=7)[/tex]:
[tex]\cos\angle BAD=\frac{64+25-49}{2 \times 8 \times 5}=\frac{40}{80}=\frac{1}{2}[/tex], i.e. [tex]\angle BAD=60^\circ[/tex].
Since [tex]AB \parallel CD[/tex], [tex]\angle ADC=180^\circ-60^\circ=120^\circ[/tex] and [tex]\cos\angle ADC=-\frac{1}{2}[/tex].

Solution:
By the Law of Cosines: [tex]b^2+c^2-a^2=2bc\cos\alpha[/tex].
[tex]4S=4 \times \frac{1}{2}bc\sin\alpha=2bc\sin\alpha[/tex].
Then [tex]2bc\cos\alpha=2bc\sin\alpha[/tex], i.e. [tex]\tan\alpha=1[/tex] and [tex]\alpha=45^\circ[/tex].

Solution:
[tex]c^2=4+2\sqrt{3}[/tex].
[tex]\cos\alpha=\frac{b^2+c^2-a^2}{2bc}=\frac{6+2\sqrt{3}}{2\sqrt{6}(1+\sqrt{3})}=\frac{2\sqrt{3}(\sqrt{3}+1)}{2\sqrt{6}(\sqrt{3}+1)}=\frac{\sqrt{2}}{2}[/tex], [tex]\alpha=45^\circ[/tex].
[tex]\cos\beta=\frac{a^2+c^2-b^2}{2ac}=\frac{2+2\sqrt{3}}{4(1+\sqrt{3})}=\frac{1}{2}[/tex], [tex]\beta=60^\circ[/tex].
[tex]\gamma=180^\circ-45^\circ-60^\circ=75^\circ[/tex].

Solution:
Let [tex]a=t[/tex], [tex]b=t\sqrt{3}[/tex], [tex]c=kt[/tex]. By the Law of Cosines for [tex]\beta[/tex]:
[tex]3t^2=t^2+k^2t^2-2 \times t \times kt \times \left(-\frac{1}{2}\right) \Rightarrow k^2+k-2=0[/tex].
Since [tex]k>0[/tex], [tex]k=1[/tex]. Then [tex]a=c[/tex] and [tex]\alpha=\gamma=\frac{180^\circ-120^\circ}{2}=30^\circ[/tex].

Solution:
The inscribed triangle cuts three triangles off the large one. Each has a [tex]60^\circ[/tex] angle opposite a side of length [tex]\sqrt{3}[/tex], and their angles at the vertices of the small triangle are equal in pairs (two adjacent ones add up to [tex]120^\circ[/tex]). So the three triangles are congruent, and each side of the large triangle is divided into segments [tex]x[/tex] and [tex]3-x[/tex].
By the Law of Cosines: [tex]3=x^2+(3-x)^2-2x(3-x)\cos 60^\circ=3x^2-9x+9[/tex].
[tex]x^2-3x+2=0 \Rightarrow x=1[/tex] or [tex]x=2[/tex].
The segments are [tex]1[/tex] and [tex]2[/tex].

Solution:
By the Law of Cosines in [tex]\triangle ABC[/tex]: [tex]\cos\angle ABC=\frac{9+16-36}{2 \times 3 \times 4}=-\frac{11}{24}[/tex].
If the quadrilateral were cyclic, then [tex]\angle ABC=180^\circ-30^\circ=150^\circ[/tex] and [tex]\cos\angle ABC=-\frac{\sqrt{3}}{2}[/tex].
But [tex]-\frac{11}{24} \ne -\frac{\sqrt{3}}{2}[/tex], so no circle can be circumscribed.