Find, correct to the nearest degree, the three angles of the triangle with the given vertices. P(2, 0), Q(0, 3), R(3, 4)

Question

Find, correct to the nearest degree, the three angles of the triangle with the given vertices.  P(2, 0), Q(0, 3), R(3, 4)

delilnaT · Accepted Answer

I find the solution:

Jeffrey Jordon · Accepted Answer

P(2,0) Q(0,3), R(3,4)  Find vectors to represent sides  PQ→=&amp;lt;0−2,3−0&amp;gt;=&amp;lt;−2,3&amp;gt;  PR→=&amp;lt;3−2,4−0&amp;gt;=&amp;lt;1,4&amp;gt;  QR→=&amp;lt;3−0,4−3&amp;gt;=&amp;lt;3,1&amp;gt;  Find the magnitude of the vectors,  |PQ→|=(−2)2+32=13  |PR→|=12+42=17  |QR→|=32+12=10  To find the angle P, we will use the vector PQ→ and PR→, make sure that the two vectors originate from the same point.  Use the dot product to find the angle cos⁡θ=a⋅b|a||b|→θ=arccos⁡a⋅b|a||b|  PQ→,PR→: θ=arccos⁡PQ→⋅PR→|PQ→||PR→|  =arccos⁡−2⋅1+3⋅41317=arccos⁡−2+1213⋅17≈47.73∘  f(x,y)=ye−x⇒fx(x,y)=−ye−x and fy(x,y)=e−x  If u is a unit vector in the direction θ=2π3, the from equation  Du=f(0,4)=fx(0,4)cos⁡(2π3)+fy(0,4)sin⁡(2π3)  =−4⋅−12+1⋅32=2+32  Result:  Duf(0,4)=2+32

alenahelenash · Accepted Answer

Step 1:First, we find the vector PQ→:PQ→=⟨xQ−xP,yQ−yP⟩=⟨0−2,3−0⟩=⟨−2,3⟩.Next, we find the vector PR→:PR→=⟨xR−xP,yR−yP⟩=⟨3−2,4−0⟩=⟨1,4⟩.Step 2:To calculate the angle between these two vectors, we can use the dot product formula:cos(θ)=PQ→·PR→‖PQ→‖‖PR→‖, where PQ→·PR→ is the dot product of the vectors, and ‖PQ→‖ and ‖PR→‖ are the magnitudes of the vectors.The dot product PQ→·PR→ is given by:PQ→·PR→=(−2)(1)+(3)(4)=−2+12=10.The magnitudes of the vectors are:‖PQ→‖=(−2)2+32=4+9=13,‖PR→‖=12+42=1+16=17.Substituting these values into the formula, we have:cos(θ)=101317.Step 3:To find the angle θ, we can take the inverse cosine (arccos) of the result:θ=arccos(101317).Now we can calculate θ using a calculator or computer software to get the value in degrees.

user_27qwe · Accepted Answer

To find the three angles of the triangle with vertices P(2,0), Q(0,3), and R(3,4), we can use the distance formula and the Law of Cosines. Let&#039;s begin by calculating the lengths of the sides of the triangle using the distance formula.The distance dPQ between points P and Q is given by:dPQ=(xQ−xP)2+(yQ−yP)2Substituting the coordinates of P(2,0) and Q(0,3) into the formula, we have:dPQ=(0−2)2+(3−0)2=4+9=13Similarly, we can calculate the distances dPR and dQR as follows:dPR=(xR−xP)2+(yR−yP)2dQR=(xR−xQ)2+(yR−yQ)2Substituting the coordinates of P(2,0) and R(3,4) into the formulas, we get:dPR=(3−2)2+(4−0)2=1+16=17dQR=(3−0)2+(4−3)2=9+1=10Now, let&#039;s use the Law of Cosines to find the angles of the triangle.The Law of Cosines states that for any triangle with sides of lengths a, b, and c, and the angle opposite side c denoted as θ, the following equation holds:c2=a2+b2−2abcos(θ)We can use this formula to find each angle of the triangle.Angle P is opposite side dPQ:cos(P)=dQR2+dPR2−dPQ22·dQR·dPRSubstituting the calculated values, we get:cos(P)=(10)2+(17)2−(13)22·10·17Similarly, we can find angles Q and R using the Law of Cosines.Angle Q is opposite side dQR:cos(Q)=dPR2+dPQ2−dQR22·dPR·dPQAngle R is opposite side dPR:cos(R)=dPQ2+dQR2−dPR22·dPQ·dQRNow, we can use the inverse cosine function to find the angles:Angle P:P=cos−1(cos(P))Angle Q:Q=cos−1(cos(Q))Angle R:R=cos−1(cos(R))Finally, we can calculate the angles using a calculator or software to find the angles to the nearest degree.

Find, correct to the nearest degree, the three angles of the triangle with the given vertices. P(2, 0), Q(0, 3), R(3, 4)

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