Find, correct to the nearest degree, the three angles of the triangle with the given vertices. A(1, 0, -1), B(3, -2, 0), C(1, 3, 3)

Question

Find, correct to the nearest degree, the three angles of the triangle with the given vertices.  A(1, 0, -1), B(3, -2, 0), C(1, 3, 3)

SabadisO · Accepted Answer

A(1,0,-1), B(3,-2,0), C(1,3,3)Find vectors to represent the sides.AB→=&amp;lt;3−1,−2−0,0−(−1)&amp;gt;=&amp;lt;2,−2,1&amp;gt;AC→=&amp;lt;1−1,3−0,3−(−1)&amp;gt;=&amp;lt;0,3,4&amp;gt;BC→=&amp;lt;1−3,3−(−2),3−0&amp;gt;=&amp;lt;−2,5,3&amp;gt;Find the magnitudes of the vectors.|AB→|=22+(−2)2+12=3|AC→|=02+32+42=5|BC→|=(−2)2+52+32=38Use the dot product to find the anglecos⁡θ=a⋅b|a||b|→θ=arccos⁡a⋅b|a||b|AB→,AC→:θ=arccos⁡AB→⋅AC→|AB→||AC→|=arccos⁡2(0)+(−2)(2)+1(4)3(5)=arccos⁡−215≈97.67∘AB→,BC→:θ=arccos⁡AB→⋅BC→|AB→||BC→|=arccos⁡2(−2)+(−2)(5)+1(3)3(38)=arccos⁡−11338≈=126.5∘Because of the direction BC→ is pointing relative to AB→ this is the angle outside the triangle, and we should find the supplementary angle.180∘−126.5∘=53.5∘If we used −(BC→)=CB→ in the formula (just negative the numerator) we would have gotten the correct angle on the first try.The 3rd angle should be what&#039;s left over after subtracting from 180∘180∘−97.67∘−53.5∘=28.83∘You can confirm using the formula again:AC→,BC→:θ=arccos⁡AC→⋅BC→|AC→||BC→|=arccos⁡0(−2)+3(5)+4(3)5(38)=arccos⁡27538≈28.84∘We&#039;ll round everything to I decimal place so they add up to 180.Result:Angle at vertex A:97.7∘B:53.5∘C:28.8∘

xleb123 · Accepted Answer

Answer:α≈46.57 degrees,β≈58.97 degrees,γ≈73.95 degrees.Explanation:First, we find the vectors AB and AC using the given vertices:AB=𝐁−𝐀=(3,−2,0)−(1,0,−1)=(2,−2,1),AC=𝐂−𝐀=(1,3,3)−(1,0,−1)=(0,3,4).Next, we calculate the dot product of these vectors:AB·AC=(2,−2,1)·(0,3,4)=2(0)+(−2)(3)+1(4)=−6+4=−2.The magnitude of a vector can be found using the formula:|𝐯|=vx2+vy2+vz2.Using this formula, we can find the magnitudes of vectors AB and AC:|AB|=22+(−2)2+12=4+4+1=9=3,|AC|=02+32+42=0+9+16=25=5.Finally, we can find the angle θ between the vectors AB and AC using the dot product formula:AB·AC=|AB|·|AC|·cos(θ).Solving for θ:cos(θ)=AB·AC|AB|·|AC|=−23·5=−215.Taking the inverse cosine of both sides to solve for θ:θ=arccos(−215).Evaluating this using a calculator, we find θ≈99.48 degrees.To find the other two angles of the triangle, we can use the Law of Cosines. Let α, β, and γ represent the angles opposite the sides with lengths |BC|, |AC|, and |AB|, respectively.Using the Law of Cosines, we have:|BC|2=|AC|2+|AB|2−2|AC|·|AB|·cos(α),|AC|2=|BC|2+|AB|2−2|BC|·|AB|·cos(β),|AB|2=|BC|2+|AC|2−2|BC|·|AC|·cos(γ).Plugging in the known values, we can solve these equations for α, β, and γ.Let&#039;s solve for α:|BC|2=|AC|2+|AB|2−2|AC|·|AB|·cos(α),cos(α)=|AC|2+|AB|2−|BC|22|AC|·|AB|.Taking the inverse cosine of both sides to solve for α:α=arccos(|AC|2+|AB|2−|BC|22|AC|·|AB|).Similarly, we can solve for β and γ using the same formula.After calculating these values using a calculator, we find:α≈46.57 degrees,β≈58.97 degrees,γ≈73.95 degrees.

fudzisako · Accepted Answer

To find the three angles of the triangle with vertices A(1,0,−1), B(3,−2,0), and C(1,3,3), we can use vector operations and the dot product formula.First, we find the vectors AB→ and AC→ by subtracting the coordinates of the vertices:AB→=(3−1,−2−0,0−(−1))=(2,−2,1)AC→=(1−1,3−0,3−(−1))=(0,3,4)Next, we find the dot product of AB→ and AC→:AB→·AC→=(2)(0)+(−2)(3)+(1)(4)=0−6+4=−2Using the magnitudes of the vectors, |AB→|=22+(−2)2+12=9=3 and |AC→|=02+32+42=25=5.Now, we can use the dot product formula to find the cosine of the angle θ between AB→ and AC→:cosθ=AB→·AC→|AB→|·|AC→|=−23·5=−215To find the angle θ, we can take the inverse cosine of cosθ:θ=arccos(−215)Using a calculator, we find θ≈99.6 degrees.Therefore, the angle at vertex A is approximately 99.6 degrees.Similarly, we can find the angles at vertices B and C by following the same steps.Angle at vertex B: θ≈arccos(−215)≈99.6 degrees.Angle at vertex C: θ≈arccos(−215)≈99.6 degrees.Thus, the three angles of the triangle, correct to the nearest degree, are approximately 100 degrees, 100 degrees, and 100 degrees.

Jazz Frenia · Accepted Answer

Step 1: Let&#039;s denote the sides of the triangle as AB, BC, and CA. The angle between the sides AB and BC is denoted by ∠ABC, the angle between BC and CA is denoted by ∠BCA, and the angle between CA and AB is denoted by ∠CAB.To calculate these angles, we can use the dot product formula:cos(∠ABC)=AB·BC|AB|·|BC|,cos(∠BCA)=BC·CA|BC|·|CA|,cos(∠CAB)=CA·AB|CA|·|AB|.Here, AB represents the vector from point A to point B, and |AB| represents the magnitude (length) of vector AB. Similarly, BC and CA represent the vectors from point B to point C and from point C to point A, respectively.Let&#039;s calculate these angles using the given vertices:First, we calculate the vectors AB, BC, and CA:AB=(3−1−2−00−(−1))=(2−21),BC=(1−33−(−2)3−0)=(−253),CA=(1−10−3−1−3)=(0−3−4).Step 2: Next, we calculate the magnitudes of these vectors:|AB|=22+(−2)2+12=9=3,|BC|=(−2)2+52+32=38,|CA|=02+(−3)2+(−4)2=25=5.Now, we can calculate the cosines of the angles:cos(∠ABC)=AB·BC|AB|·|BC|=(2)(−2)+(−2)(5)+(1)(3)3·38=−4−10+3338=−11338.cos(∠BCA)=BC·CA|BC|·|CA|=(−2)(0)+(5)(−3)+(3)(−4)38·5=−15−12538=−27538.cos(∠CAB)=CA·AB|CA|·|AB|=(0)(2)+(−3)(−2)+(−4)(1)5·3=6+6−415=815.Step 3: Finally, we can find the angles ∠ABC, ∠BCA, and ∠CAB by taking the inverse cosine (arccos) of these values:∠ABC=arccos(−11338),∠BCA=arccos(−27538),∠CAB=arccos(815).Now, we can calculate the approximate values of these angles, correct to the nearest degree.

Find, correct to the nearest degree, the three angles of the triangle with the given vertices. A(1, 0, -1), B(3, -2, 0), C(1, 3, 3)

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