Triangle

Triangle KLM is given by plane coordinates of vertices: K[11, -10] L[10, 12] M[1, 3].

Calculate its area and itsinterior angles.

Result

S =  103.5
K =  34.966 °
L =  47.603 °
M =  97.431 °

Solution:

x0=11 y0=10  x1=10 y1=12  x2=1 y2=3   LM=ML=(k0,k1) k0=x2x1=110=9 k1=y2y1=312=9  KM=MK=(l0,l1) l0=x2x0=111=10 l1=y2y0=3(10)=13  LK=KL=(m0,m1) m0=x0x1=1110=1 m1=y0y1=(10)12=22   k=k02+k12=(9)2+(9)2=9 212.7279 l=l02+l12=(10)2+132=26916.4012 m=m02+m12=12+(22)2=48522.0227  s=k+l+m2=12.7279+16.4012+22.0227225.5759 S=s (sk) (sl) (sm)=25.5759 (25.575912.7279) (25.575916.4012) (25.575922.0227)=2072=103.5x_{ 0 } = 11 \ \\ y_{ 0 } = -10 \ \\ \ \\ x_{ 1 } = 10 \ \\ y_{ 1 } = 12 \ \\ \ \\ x_{ 2 } = 1 \ \\ y_{ 2 } = 3 \ \\ \ \\ \ \\ LM = M-L = (k_{ 0 },k_{ 1 }) \ \\ k_{ 0 } = x_{ 2 }-x_{ 1 } = 1-10 = -9 \ \\ k_{ 1 } = y_{ 2 }-y_{ 1 } = 3-12 = -9 \ \\ \ \\ KM = M-K = (l_{ 0 },l_{ 1 }) \ \\ l_{ 0 } = x_{ 2 }-x_{ 0 } = 1-11 = -10 \ \\ l_{ 1 } = y_{ 2 }-y_{ 0 } = 3-(-10) = 13 \ \\ \ \\ LK = K-L = (m_{ 0 },m_{ 1 }) \ \\ m_{ 0 } = x_{ 0 }-x_{ 1 } = 11-10 = 1 \ \\ m_{ 1 } = y_{ 0 }-y_{ 1 } = (-10)-12 = -22 \ \\ \ \\ \ \\ k = \sqrt{ k_{ 0 }^2+k_{ 1 }^2 } = \sqrt{ (-9)^2+(-9)^2 } = 9 \ \sqrt{ 2 } \doteq 12.7279 \ \\ l = \sqrt{ l_{ 0 }^2+l_{ 1 }^2 } = \sqrt{ (-10)^2+13^2 } = \sqrt{ 269 } \doteq 16.4012 \ \\ m = \sqrt{ m_{ 0 }^2+m_{ 1 }^2 } = \sqrt{ 1^2+(-22)^2 } = \sqrt{ 485 } \doteq 22.0227 \ \\ \ \\ s = \dfrac{ k+l+m }{ 2 } = \dfrac{ 12.7279+16.4012+22.0227 }{ 2 } \doteq 25.5759 \ \\ S = \sqrt{ s \cdot \ (s-k) \cdot \ (s-l) \cdot \ (s-m) } = \sqrt{ 25.5759 \cdot \ (25.5759-12.7279) \cdot \ (25.5759-16.4012) \cdot \ (25.5759-22.0227) } = \dfrac{ 207 }{ 2 } = 103.5
K=angle(KL,KM) K1=arccos(l0 m0l1 m1l m)=arccos((10) 113 (22)16.4012 22.0227)0.6103 K=K1 =K1 180π  =34.9660298264  =34.966=345758"K = angle(KL, KM) \ \\ K_{ 1 } = \arccos(\dfrac{ -l_{ 0 } \cdot \ m_{ 0 }-l_{ 1 } \cdot \ m_{ 1 } }{ l \cdot \ m } ) = \arccos(\dfrac{ -(-10) \cdot \ 1-13 \cdot \ (-22) }{ 16.4012 \cdot \ 22.0227 } ) \doteq 0.6103 \ \\ K = K_{ 1 } \rightarrow \ ^\circ = K_{ 1 } \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ = 34.9660298264 \ \ ^\circ = 34.966 ^\circ = 34^\circ 57'58"
L=angle(LK,LM) L1=arccos(k0 m0+k1 m1k m)=arccos((9) 1+(9) (22)12.7279 22.0227)0.8308 L=L1 =L1 180π  =47.6025622026  =47.603=47369"L = angle(LK, LM) \ \\ L_{ 1 } = \arccos(\dfrac{ k_{ 0 } \cdot \ m_{ 0 }+k_{ 1 } \cdot \ m_{ 1 } }{ k \cdot \ m } ) = \arccos(\dfrac{ (-9) \cdot \ 1+(-9) \cdot \ (-22) }{ 12.7279 \cdot \ 22.0227 } ) \doteq 0.8308 \ \\ L = L_{ 1 } \rightarrow \ ^\circ = L_{ 1 } \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ = 47.6025622026 \ \ ^\circ = 47.603 ^\circ = 47^\circ 36'9"
M1=arccos(k0 l0+k1 l1k l)=arccos((9) (10)+(9) 1312.7279 16.4012)1.7005 M=M1 =M1 180π  =97.4314079711  =97.431=972553"M_{ 1 } = \arccos(\dfrac{ k_{ 0 } \cdot \ l_{ 0 }+k_{ 1 } \cdot \ l_{ 1 } }{ k \cdot \ l } ) = \arccos(\dfrac{ (-9) \cdot \ (-10)+(-9) \cdot \ 13 }{ 12.7279 \cdot \ 16.4012 } ) \doteq 1.7005 \ \\ M = M_{ 1 } \rightarrow \ ^\circ = M_{ 1 } \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ = 97.4314079711 \ \ ^\circ = 97.431 ^\circ = 97^\circ 25'53"

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For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc. Two vectors given by its magnitudes and by included angle can be added by our vector sum calculator. Cosine rule uses trigonometric SAS triangle calculator. See also our trigonometric triangle calculator. Pythagorean theorem is the base for the right triangle calculator.

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