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Home
Practice
Law of Cosines
Easy
Normal
Law of Cosines: Problems with Solutions
Problem 1
Find b = ?
3
$3\sqrt{7}$
$2\sqrt{7}$
$3\sqrt{5}$
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 $b=\sqrt{28}=2\sqrt{7}$
Problem 2
What is measure of $\alpha$?
$50.95^{\circ }$
$41.79^{\circ }$
$87.26^{\circ }$
$55.26^{\circ }$
Solution:
The Law of Cosines says $BC^{2}=AC^{2}+AB^{2}-2(AC)(AB)\cos \alpha $
then $\cos \alpha =\frac{AC^{2}+AB^{2}-BC^{2}}{2(AC)(AB)}=\frac{36+81-49}{2(6)(9)}=\frac{68}{108}=0.630$ so
$\alpha =\arccos (0.630)=0.88924(\frac{180}{\pi })= 50.95^{\circ }$
Problem 3
In a triangle
ABC
,
AC=3
,
BC=5
,
AB=6
. Find [tex]\cos(\angle ACB)[/tex].
$-\frac{1}{20}$
$-\frac{1}{15}$
$-\frac{1}{5}$
$\frac{1}{20}$
Solution:
By the law of cosines for triangle
ABC
, 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]
Problem 4
Given a triangle
ABC
with
AC=12
,
BC=10
and [tex]\angle ACB = 60^\circ [/tex], find the value of
AB
2
= ?
.
Solution:
By the law of cosines, we have [tex]AB^2=AC^2+BC^2-2.AC.BC.\cos\angle ACB[/tex]. Substituting
AC
,
BC
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].
Problem 5
Given a triangle
ABC
with
AC=17
,
BC=14
and [tex]\angle ACB = 60^\circ [/tex], find the value of
AB
2
= ?
.
Solution:
By the law of cosines, we have [tex]AB^2=AC^2+BC^2-2\cdot AC\cdot BC\cdot \cos\angle ACB[/tex]. Substituting
AC
,
BC
and the angle with their values, we get: [tex]AB^2=17^2+14^2-2\cdot 17\cdot 14\cdot \cos(60^{\circ})[/tex], or [tex]AB^2=289+196-2\cdot 238\cdot \frac{1}{2}[/tex]. By performing the arithmetical operations on the right side, we get [tex]AB^2=247[/tex].
Problem 6
Given a triangle
ABC
with
AC=22
,
BC=21
and [tex]\angle ACB = 60^\circ [/tex], find the value of
AB
2
= ?
.
Solution:
By the law of cosines, we have [tex]AB^2=AC^2+BC^2-2\cdot AC\cdot BC\cdot \cos\angle ACB[/tex]. Substituting
AC
,
BC
and the angle with their values, we get: [tex]AB^2=22^2+21^2-2\cdot 22\cdot 21\cdot \cos(60^{\circ})[/tex], or [tex]AB^2=484+441-2\cdot 462\cdot \frac{1}{2}[/tex]. By performing the arithmetical operations on the right side, we get [tex]AB^2=463[/tex].
Problem 7
Given a triangle
ABC
with
AC=8
,
BC=6
and [tex]\angle ACB = 60^\circ [/tex], find the value of
AB
2
= ?
.
Solution:
By the law of cosines, we have [tex]AB^2=AC^2+BC^2-2\cdot AC\cdot BC\cdot \cos\angle ACB[/tex]. Substituting
AC
,
BC
and the angle with their values, we get: [tex]AB^2=8^2+6^2-2\cdot 8\cdot 6\cdot \cos(60^{\circ})[/tex], or [tex]AB^2=64+36-2\cdot 48\cdot \frac{1}{2}[/tex]. By performing the arithmetical operations on the right side, we get [tex]AB^2=52[/tex].
Easy
Normal
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