Question

A triangle is defined by the three points: A=(7,9) B=(8,8) C=(6,6). How to determine all three angles in the triangle (in radians) ?

Answer 1

`A=63.435^\circ`, `B=90^\circ`, `C=26.565^\circ`

Explanation:

The lengths of sides of triangle having vertices `A(7, 9)`, `B(8,8)` & `C(6, 6)` are given as

`AB=\sqrt{(7-8)^2+(9-8)^2}=\sqrt2`

`BC=\sqrt{(8-6)^2+(8-6)^2}=2\sqrt2`

`AC=\sqrt{(7-6)^2+(9-6)^2}=\sqrt{10}`

Now, using Cosine rule in `\Delta ABC` as follows

`\cos A=\frac{AB^2+AC^2-BC^2}{2(AB)(AC)}`

`\cos A=\frac{(\sqrt2)^2+(\sqrt10)^2-(2\sqrt2)^2}{2(\sqrt2)(\sqrt10)}`

`\cos A=1/\sqrt5`

`A=\cos^{-1}(1/\sqrt5)=63.435^\circ`

Similarly, we get

`\cos B=\frac{AB^2+BC^2-AC^2}{2(AB)(BC)}`

`\cos B=\frac{(\sqrt2)^2+(2\sqrt2)^2-(\sqrt10)^2}{2(\sqrt2)(2\sqrt2)}`

`\cos B=0`

`B=90^\circ`

`\cos C=\frac{AC^2+BC^2-AB^2}{2(AC)(BC)}`

`\cos C=\frac{(\sqrt10)^2+(2\sqrt2)^2-(\sqrt2)^2}{2(\sqrt10)(2\sqrt2)}`

`\cos C=2/\sqrt5`

`C=\cos^{-1}(2/\sqrt5)=26.565^\circ`