Find, correct to the nearest degree, the three angles of the triangle with the given vertices. p(1, 0), q(0, 1), r(4, 4)

2 answers
Question:

Find, correct to the nearest degree, the three angles of the triangle with the given vertices. p(1, 0), q(0, 1), r(4, 4)

Answers

The angles of the triangle are 81.870°, 81.870° and 16.260°, respectively.We can determine the angles of the triangle by using the definition of dot product as follows:[tex]\alpha = \cos^{-1} \frac{\overrightarrow {QP}\,\bullet \,\overrightarrow {QR} }{\|QP\|\cdot \|QR\|}[/tex] (1)[tex]\beta = \cos^{-1} \frac{\overrightarrow {RQ}\,\bullet \,\overrightarrow {RP} }{\|RQ\|\cdot \|RP\|}[/tex] (2)[tex]\gamma = \cos^{-1} \frac{\overrightarrow{PR}\,\bullet\,\overrightarrow {PQ}}{\|\overrightarrow {PR}\|\cdot \|\overrightarrow{PQ}\|}[/tex] (3)Where:[tex]\alpha[/tex], [tex]\beta[/tex], [tex]\gamma[/tex] - Angles of the triangle.[tex]\overrightarrow {PQ}[/tex], [tex]\overrightarrow {QR}[/tex], [tex]\overrightarrow {PR}[/tex] - Vectors of the triangle.[tex]\|\overrightarrow{PQ}\|[/tex], [tex]\|\overrightarrow{QR}\|[/tex], [tex]\|\overrightarrow{PR}\|[/tex] - Norm of the vectors of the triangle.Please remember that Norm is determined by the Pythagorean Theorem.If we know that [tex]P(x,y) = (1,0)[/tex], [tex]Q (x,y) = (0,1)[/tex] and [tex]R(x,y) = (4, 4)[/tex], then the angles of the triangle are, respectively:[tex]\overrightarrow {PQ} = (0,1)-(1,0)[/tex][tex]\overrightarrow{PQ} = (-1, 1)[/tex][tex]\overrightarrow{QP} = (1, -1)[/tex][tex]\|\overrightarrow {PQ}\| = \|\overrightarrow{QP}\| = \sqrt{2}[/tex][tex]\overrightarrow {QR} = (4,4) - (0,1)[/tex][tex]\overrightarrow {QR} = (4, 3)[/tex][tex]\overrightarrow {RQ} = (-4, -3)[/tex][tex]\|\overrightarrow{QR}\| = \|\overrightarrow {RQ}\| = 5[/tex][tex]\overrightarrow {PR} = (4,4) - (1,0)[/tex][tex]\overrightarrow {PR} = (3,4)[/tex][tex]\overrightarrow{RP} = (-3,-4)[/tex][tex]\|\overrightarrow {PR}\| = \|\overrightarrow{RP}\| = 5[/tex][tex]\alpha = \cos^{-1} \frac{(1,-1)\,\bullet (4,3)}{5\sqrt{2}}[/tex][tex]\alpha \approx 81.870^{\circ}[/tex][tex]\beta = \cos^{-1} \frac{(-4,-3)\,\bullet\,(-3, -4)}{25}[/tex][tex]\beta \approx 16.260^{\circ}[/tex][tex]\gamma = \cos^{-1} \frac{(3,4)\,\bullet\,(-1, 1)}{5\sqrt{2}}[/tex][tex]\gamma \approx 81.870^{\circ}[/tex]The angles of the triangle are 81.870°, 81.870° and 16.260°, respectively.We kindly invite to check this question on triangles: https://brainly.com/question/15161714

A geometry app shows the angles to be  ∠p ≈ 82°  ∠q ≈ 82°  ∠r ≈ 16°_____The length of pq is √2, and the length of pr is 5. Then   cos(p) = √2/(2·5)  p = arccos(√2/10) = 81.87°The triangle is isosceles, so q = 81.87° and r = 180° -2*81.87° = 16.26°.

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