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) ?

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Answer 1 · 2018-06-28T22:47:57

$A = {63.435}^{\circ}$ $A=63.435^\circ$ , $B = 90^{\circ}$ $B=90^\circ$ , $C = {26.565}^{\circ}$ $C=26.565^\circ$

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

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

$B C = \sqrt{{(8 - 6)}^{2} + {(8 - 6)}^{2}} = 2 \sqrt{2}$ $BC=\sqrt{(8-6)^2+(8-6)^2}=2\sqrt2$

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

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

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

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

$\cos B=0$

$B=90^\circ$

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

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

$\cos C=2/\sqrt5$

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