In trigonometry the cosine rule relates the lengths of the sides and the cosine of one of its angles.
The Law of Cosines is useful in finding:
ABC is a triangle with side lengths $BC=a$BC=a , $CA=b$CA=b and $AB=c$AB=c and the opposite angles of the sides are respectively angle $A$A, angle $B$B and angle $C$C.
$a^2=b^2+c^2-2bc\cos A$a2=b2+c2−2bccosA
$b^2=a^2+c^2-2ac\cos B$b2=a2+c2−2accosB
$c^2=a^2+b^2-2ab\cos C$c2=a2+b2−2abcosC
Notice that Pythagoras' Theorem $a^2=b^2+c^2$a2=b2+c2 makes an appearance in the Cosine Rule: $a^2=b^2+c^2-2bc\cos A$a2=b2+c2−2bccosA
Find angle $B$B in the triangle.
Think: All three side lengths are known, so I can apply the cosine rule. The unknown angle B appears opposite side $b=3$b=3.
$b^2$b2 | $=$= | $a^2+c^2-2ac\cos B$a2+c2−2accosB |
$3^2$32 | $=$= | $5^2+6^2-2\times5\times6\cos B$52+62−2×5×6cosB |
$9$9 | $=$= | $25+36-60\cos B$25+36−60cosB |
$9-61$9−61 | $=$= | $-60\cos B$−60cosB |
$\frac{-52}{-60}$−52−60 | $=$= | $\cos B$cosB |
$\cos B$cosB | $=$= | $0.866667$0.866667 |
$B$B | $=$= | $29.9^\circ$29.9° to $1$1 decimal place |
Find the value of $x$x in the diagram.
Think: The first thing I always do is identify which side is opposite the given angle. This side is the subject of the formula. To find out which other values we are given I label the sides and angles using $a$a,$b$b and $c$c .
Do:
I add the following labels to the triangle:
So I want to find the value of $c$c.
$c^2$c2 | $=$= | $a^2+b^2-2ab\cos C$a2+b2−2abcosC |
$c^2$c2 | $=$= | $8^2+11^2-2\times8\times11\cos39^\circ$82+112−2×8×11cos39° |
$c^2$c2 | $=$= | $64+121-176\cos39^\circ$64+121−176cos39° |
$c^2$c2 | $=$= | $48.22$48.22 |
$c$c | $=$= | $6.94$6.94 |
The following interactive demonstrates that the cosine rule holds regardless of the angles or size and shape of the triangle.
Find the length of $a$a using the cosine rule.
Round your answer to two decimal places.
Find the length of $c$c using the cosine rule.
Round your answer to two decimal places.