Question

There are three islands that form a triangle. Island A is 6 miles from island B and 5 miles from island C. Island B is 7 miles from island C. Find the three angles (to the nearest degree) that are created by these three islands.

Answer 1

The information can be represented in a diagram as follows

Island B 6miles 7miles 5miles Island C Island A

Island A is 6 miles from Island B, and 5 miles from Island B. Island B is 7 miles from Island C. Let the angle produced by Islands A,B and C be A,B and C respectively

The three triangles form a triangle

As per the law of cosines, if a,b,c are the 3 sides of a triangle, and A,B,C are angles such that a,b,c are respectivelly opposite them, then

$Cos(A)=(b^2+c^2-a^2)/2bc$

$Cos(B)=(a^2+c^2-b^2)/2ac$

Cos(C) = (a? +62 - c?)/2ab

Hence, in this case,

$a=7,b=5,c=6$

Applying law of cosines to Angle A, formed by Island A, we get

$Cos(A)=(5^2+6^2-7^2)/2*6*5$

$Cos(A)=(25+36-49)/60$

$Cos(A)=12/60=0.2$

$A=cos^{-1}(0.2)$

$A=78.46^0$

Similarly, applying law of cosines to angle B, we get

$Cos(B)=(a^2+c^2-b^2)/2ac$
$Cos(B)=(7^2+6^2-5^2)/2*7*6$

$Cos(B)=(49+36-25)/84$

$Cos(B)=60/84$

$Cos(B)=0.714285714$

$(B)=cos^{-1}(0.714285714)$

$(B)=44.415^0$

Also, we know that the angles of a triangle add up to $180^0$

$<A+<B+<C=180^0$

$78.46^0+44.415^0+<C=180^0$

$122.875^0+<C=180^0$

$<C=180^0-122.875^0$

$<C=57.125^0$

Hence, the angles formed by the three islands are as above