Triangle

Plane coordinates of vertices: K[11, -10] L[10, 12] M[1, 3] give Triangle KLM.

Calculate its area and its interior angles.

Correct result:

S =  103.5
K =  34.966 °
L =  47.6026 °
M =  97.4314 °

Solution:

$$\begin{gathered} x_0=11 \\ {y}_0=-10 \\ {x}_1=10 \\ {y}_1=12 \\ {x}_2=1 \\ {y}_2=3 \\ LM=M-L=(k_0,k_1) \\ {k}_0=x_2-x_1=1-10=-9 \\ {k}_1=y_2-y_1=3-12=-9 \\ KM=M-K=(l_0,l_1) \\ {l}_0=x_2-x_0=1-11=-10 \\ {l}_1=y_2-y_0=3-(-10)=13 \\ LK=K-L=(m_0,m_1) \\ {m}_0=x_0-x_1=11-10=1 \\ {m}_1=y_0-y_1=(-10)-12=-22 \\ k=\sqrt{k_0^2+k_1^2}=\sqrt{(-9{)}^2+(-9{)}^2}=9\sqrt{2}\doteq 12.7279 \\ l=\sqrt{l_0^2+l_1^2}=\sqrt{(-10{)}^2+13^2}=\sqrt{269}\doteq 16.4012 \\ m=\sqrt{m_0^2+m_1^2}=\sqrt{1^2+(-22{)}^2}=\sqrt{485}\doteq 22.0227 \\ s={\displaystyle \frac{k+l+m}{2}}={\displaystyle \frac{12.7279+16.4012+22.0227}{2}}\doteq 25.5759 \\ S=\sqrt{s\cdot (s-k)\cdot (s-l)\cdot (s-m)}=\sqrt{25.5759\cdot (25.5759-12.7279)\cdot (25.5759-16.4012)\cdot (25.5759-22.0227)}={\displaystyle \frac{207}{2}}=103\frac{1}{2}=103.5 \end{gathered}$$

$$\begin{gathered} K=angle(KL,KM) \\ {K}_1=\arccos ({\displaystyle \frac{-l_0\cdot m_0-l_1\cdot m_1}{l\cdot m}})=\arccos ({\displaystyle \frac{-(-10)\cdot 1-13\cdot (-22)}{16.4012\cdot 22.0227}})\doteq 0.6103 \\ K=K_1{\to }^{\circ }=K_1\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=0.6103\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=34.966^{\circ }=34^{\circ }57^{\mathrm{\prime }}58\mathrm{"} \end{gathered}$$

$$\begin{gathered} L=angle(LK,LM) \\ {L}_1=\arccos ({\displaystyle \frac{k_0\cdot m_0+k_1\cdot m_1}{k\cdot m}})=\arccos ({\displaystyle \frac{(-9)\cdot 1+(-9)\cdot (-22)}{12.7279\cdot 22.0227}})\doteq 0.8308 \\ L=L_1{\to }^{\circ }=L_1\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=0.8308\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=47.603^{\circ }=47.6026^{\circ }=47^{\circ }36^{\mathrm{\prime }}9\mathrm{"} \end{gathered}$$

$$\begin{gathered} M_1=\arccos ({\displaystyle \frac{k_0\cdot l_0+k_1\cdot l_1}{k\cdot l}})=\arccos ({\displaystyle \frac{(-9)\cdot (-10)+(-9)\cdot 13}{12.7279\cdot 16.4012}})\doteq 1.7005 \\ M=M_1{\to }^{\circ }=M_1\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=1.7005\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=97.431^{\circ }=97.4314^{\circ }=97^{\circ }25^{\mathrm{\prime }}53\mathrm{"} \end{gathered}$$

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Showing 5 comments:

Math student

It's Great!. Am grateful.

3 years ago  1 Like Like

Math student

Still don't get it though

2 years ago  Like

Math student

I find it hard ... But I think I will get there. ... Slowly but surely ...

1 year ago  Like

Math student

I still dont understand

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I need one question

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