# Cosine Law

Cosine law is a formula used to find the third of SAS triangle. A triangle with two sides given and an included angle. Solving problems that requires law of cosine as solution is easy to recognize.

Derivation:

This derivation of cosine law is graphical method. Given an oblique triangle and put that triangle inside a rectangular plane. Given a triangle ABC above where side AB is on positive x-axis. By dropping a perpendicular line trough C we create triangle ACD which is a right triangle.

Let y be the distance from point C to D. Let x be the distance from point A to D.

Once we hear a right triangle in trigonometry, that is always associated with SOH-CAH-TOA formula.

In $\triangle ACD$ we have the following relations. $sinA=\displaystyle\frac{y}{b}$ $cosA=\displaystyle\frac{x}{b}$

Rearranging both equations we have, $y=bsinA$ $x=bcosA$

Observe that (x,y) is the coordinate of C. Thus, (x,y)=(bcosA,bsinA).

Also the coordinate of B is (c,0).

Applying the distance formula between B and C we have, $d^2=(x_2-x_1)^2+(y_2-y_1)^2$ $a^2=(bcosA-c)^2+(bsinA-0)^2$ $a^2=b^2cos^2A-2bccosA+c^2+b^2sin^2A$

Rearranging the expression we have, $a^2=b^2cos^2A+b^2sin^2A+c^2-2bccosA$

By factoring the common factor $b^2$ $a^2=b(cos^2A+sin^2A)+c^2-2bccosA$

But $cos^2A+sin^2A=1$ $a^2=b^2+c^2-2bccosA$

Given a triangle below,  $b^2=a^2+c^2-2accosB$ $a^2=b^2+c^2-2bccosA$ $c^2=a^2+b^2-2abcosC$

Consider if the triangle is a right triangle where C=90°, $c^2=a^2+b^2-2abcosC$ $c^2=a^2+b^2-2abcos90$

cos90°=0 $c^2=a^2+b^2-2ab(0)$ $c^2=a^2+b^2$

This reduces to the famous Pythagorean Theorem.

If the SAS triangle ( triangle with two adjacent side and included angle) is given and ask for the third side. Kill the problem using cosine law.

There are problems that you need to know the basic facts about geometry to unveil the hidden given.

Facts:

“Radii of the same circle are congruent”

“Each side of regular polygon is equal”

Worked Problem 1:

Find the length of each side of regular octagon if the length of the longest diagonal is 10 cm.

Solution: The longest diagonal of a polygon is drawn from one vertex to another vertex passing the center of the given polygon. Why? ( there is a circle lurking around the octagon and the longest diagonal is the diameter of the circle ) $8\theta=360$ $\theta=45$

Using cosine law, $s^2=5^2+5^2-2(5)(5)cos(45)$ $s^2=25(2-\sqrt{2})$ $s=5\sqrt{2-\sqrt{2}}=3.83$ cm

Worked Problem 2:

Find the length of the rectangle if the length of the diagonal is 20 cm and the diagonal intersects at an angle of 60°.

Solution: The side opposite of the larger angle is also the longer side. Hence, the side opposite 120° is the length of the rectangle.

Using cosine law, $L^2=10^2+10^2-2(10)(10)cos120$ $L^2=300$ $L=\sqrt{300}$ $L=10\sqrt{3}$ cm

Sample Problem 3: MMC 2007 Elimination Round

Two points M and N are on the opposite sides of a river. C is point on the same side of the river such that $\angle{MCN}=60$°. If $MC=52 m$ and $CN=72 m$. How far M is from N?

Solution: Using cosine law, $\overline{MN}^2=52^2+72^2-2(52)(72)cos60$ $\overline{MN}^2=4144$ $\overline{MN}=4\sqrt{259} m$ $\Delta ABC$ is a right triangle with right angle at $C$. $D$ is on $\overline{AB}$ such that segment $\overline{CD}$ divides $\Delta ABC$ into two triangles of equal perimeter. If $\overline{BC}=15$ and $\overline{AC}=8$, how long is $\overline{CD}$?

Solution: By Pythagorean Theorem, $AB^2=AC^2+BC^2$ $AB^2=8^2+15^2$ $AB^2=289$ $AB=17$

Let $AD=a$ then, $BD=17-a$. $Perimeter_{\Delta ADC}=Perimeter_{\Delta DCB}$ $8+x+a=15+x+17-a$ $2a=24$ $a=12$

Also, in $\Delta ABC$ $cosA=\displaystyle\frac{Adjacent}{Hypotenuse}$ $cosA=\displaystyle\frac{8}{17}$

In $\Delta ACD$, $x^2=a^2+8^2-2(a)(8)cosA$ $x^2=12^2+8^2-2(12)(8)\displaystyle\frac{8}{17}$ $x^2=\displaystyle\frac{2000}{17}$ $x=\sqrt{\displaystyle\frac{2000}{17}}$ $x=\displaystyle\frac{20\sqrt{85}}{17}$