Best solutions to - "A triangle is defined by the three points:..."

Cherish Sellers

Answered question

2022-02-05

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)?

skullsxtest7xt

Beginner2022-02-06Added 15 answers

Step 1
The lengths of sides of triangle having vertices

A (7, 9), B (8, 8)

and

C (6, 6)

are given as

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

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

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

Now, using Cosine rule in

△ A B C

as follows

\cos A = \frac{A B^{2} + A C^{2} - B C^{2}}{2 (A B) (A C)}

\cos A = \frac{{(\sqrt{2})}^{2} + {(\sqrt{10})}^{2} - {(2 \sqrt{2})}^{2}}{2 (\sqrt{2}) (\sqrt{10})}

\cos A = \frac{1}{\sqrt{5}}

A = \cos^{- 1} (\frac{1}{\sqrt{5}}) = {63.435}^{\circ}

Similarly, we get

\cos B = \frac{A B^{2} + B C^{2} - A C^{2}}{2 (A B) (B C)}

\cos B = \frac{{(\sqrt{2})}^{2} + {(2 \sqrt{2})}^{2} - {(\sqrt{10})}^{2}}{2 (\sqrt{2}) (2 \sqrt{2})}

\cos B = 0

B = 90^{\circ}

\cos C = \frac{A C^{2} + B C^{2} - A B^{2}}{2 (A C) (B C)}

\cos C = \frac{{(\sqrt{10})}^{2} + {(2 \sqrt{2})}^{2} - {(\sqrt{2})}^{2}}{2 (\sqrt{10}) (2 \sqrt{2})}

\cos C = \frac{2}{\sqrt{5}}

C = \cos^{- 1} (\frac{2}{\sqrt{}}

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