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,

Cosie Law

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:

cosinelaw2

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:

cosinelaw3

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:

cosinelaw4

Using cosine law,

\overline{MN}^2=52^2+72^2-2(52)(72)cos60

\overline{MN}^2=4144

\overline{MN}=4\sqrt{259} m

Sample Problem 4: Adapted from Canadian Open Mathematics Challenge

\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:

cosinelaw5

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}

Dan

Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
Dan

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