# Cosine Rule

• Feb 4th 2009, 09:22 AM
berniecole
Cosine Rule
I need to calculate the permeter of a triangle ABC
Side AB 9cm, Side BC 15cm and Angle B 110°

Using the Cosine Rule I Know
b² = a²+c²-2accos B
b² = 15²+9²-2x15x9xcos110°
It is claimed that
b²=398.345
b=20cm approx, a scaled drawing proves this to be the case.

My problem is, I want to make it

b² = 15²+9²-2x15x9xcos110°
b² = 225+81-2x15x9x.3420
b² = 225+81-92.34
b² = 306-92.34
b² = 213.66
b = 14.62 approx which is wrong!!

If I add 92.34 to 306 I get the correct answer but I do not know where I have gone wrong. Can anybody help me?

berniecole
• Feb 4th 2009, 10:37 AM
Soroban
Hello, berniecole!

It looks like you aren't using your calculator well . . .

Quote:

Calculate the perimeter of a triangle ABC:
$c= 9,\;a = 15,\;\angle B = 110^o$

Using the Cosine Rule:

. $b^2\:=\:a^2+c^2-2ac\cos B \;=\;15^2+9^2-2(15)(9)\cos110^o \;=\; 398.3454387$

. . $b\:=\:19.9585931 \:\approx\:20$

My problem is, I want to make it

. . $b^2 \;=\;15^2+9^2-2(15)(9)\cos110^o$ .
. . . I don't see any difference

. . . . $=\;225+81-(2)(15)(9)\underbrace{({\color{red}0.3420})}_{\tex t{No!}}$

$\cos110^o \;=\;{\color{red}-}0.342020143$

$110^o$ is in Quadrant 2, where cosine is negative.