• Jun 2nd 2008, 11:30 AM
sa87uk
Four rods, AB 20CM long, BC 22.5CM long, CD 25CM long and DA 27.5 long,
are pin-jointed at ABC and D. If the angle ABC is 120°, find

a. the distance AC

• Jun 2nd 2008, 12:56 PM
sean.1986
a)

Using cosine rule...

AC^2 = 20^2 + 22.5^2 - 2*20*22.5Cos120 = 1356.25

AC = 36.8 cm

b)

Again using cosine rule...

36.8^2 = 27.5^2 + 25^2 - 2*27.5*25CosADC

2*27.5*25CosADC = 27.5^2 + 25^2 - 36.8^2

ADC = Cos^-1((27.5^2 + 25^2 - 36.8^2)/(2*27.5*25)) = 90.0 degrees
• Jun 2nd 2008, 01:52 PM
Soroban
Hello, sa87uk!

Quote:

Four rods: AB 20cm long, BC 22.5cm long, CD 25cm long, and DA 27.5 cm long,
are jointed at A, B, C, D.

If the angle ABC is 120°, find

a) the distance AC

Code:

                22.5           B o  *  *  *  o C           * 120°    *    *           *      *        *       20 *    *              * 25         *  *                  *       * *                        *     A o  *  *  *  *  *  *  *  *  *  o D                   27.5

Law of Cosines: .$\displaystyle AC^2 \;=\;AB^2 + BC^2 - 2(AB)(BC)\cos B$

. . $\displaystyle AC^2 \;=\;20^2 + 22.5^2 - 2(20)(22.5)\cos120^o \;=\;1356.25$

(a) Therefore: .$\displaystyle AC \;\;=36.82729966 \;\approx\;36.8$ cm

Law of Cosines: .$\displaystyle \cos(\angle ADC) \;=\;\frac{AD^2 + CD^2 - AC^2}{2(AD)(CD)}$

.$\displaystyle \cos(\angle ADC) \;=\;\frac{27.5^2 + 25^2 - 1356.25}{2(27.5)(25)} \;=\;\frac{1}{55}$

(b) Therefore: .$\displaystyle \angle ADC \;=\;88.95820115^o \;\approx\;89^o$

Edit: Too fast for me, Sean!