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Question Number 42942 by Raj Singh last updated on 05/Sep/18

Answered by MJS last updated on 05/Sep/18

BC=a  CA=b  AB=c    1. a=6  β=50°  b+c=8 ⇒ c=8−b  α+γ=130° ⇒ γ=130°−α  36=b^2 +(8−b)^2 −2b(8−b)cos α  b^2 =36+(8−b)^2 −12(8−b)cos 50°  (8−b)^2 =36+b^2 −12bcos (130°−α)    ⇒ a=6  b=4.62  c=3.38        α=95.93°  β=50°  γ=34.07°    2. b=6  β=50° b+c=8 ⇒ c=2  α+γ=130° ⇒ γ=130°−α  a^2 =36+4−24cos α  36=a^2 +4−4acos 50°  4=a^2 +36−12acos (130°−α)    ⇒ a=7.09  b=6  c=2        α=115.21°  β=50°  γ=14.79°    3. c=6 β=50°  b+c=8 ⇒ b=2  α+γ=130° ⇒ γ=130°−α  a^2 =4+36−24cos α  4=a^2 +36−12acos 50°  36=a^2 +4−4acos (130°−α)    ⇒ no real solution

$${BC}={a}\:\:{CA}={b}\:\:{AB}={c} \\ $$$$ \\ $$$$\mathrm{1}.\:{a}=\mathrm{6}\:\:\beta=\mathrm{50}°\:\:{b}+{c}=\mathrm{8}\:\Rightarrow\:{c}=\mathrm{8}−{b} \\ $$$$\alpha+\gamma=\mathrm{130}°\:\Rightarrow\:\gamma=\mathrm{130}°−\alpha \\ $$$$\mathrm{36}={b}^{\mathrm{2}} +\left(\mathrm{8}−{b}\right)^{\mathrm{2}} −\mathrm{2}{b}\left(\mathrm{8}−{b}\right)\mathrm{cos}\:\alpha \\ $$$${b}^{\mathrm{2}} =\mathrm{36}+\left(\mathrm{8}−{b}\right)^{\mathrm{2}} −\mathrm{12}\left(\mathrm{8}−{b}\right)\mathrm{cos}\:\mathrm{50}° \\ $$$$\left(\mathrm{8}−{b}\right)^{\mathrm{2}} =\mathrm{36}+{b}^{\mathrm{2}} −\mathrm{12}{b}\mathrm{cos}\:\left(\mathrm{130}°−\alpha\right) \\ $$$$ \\ $$$$\Rightarrow\:{a}=\mathrm{6}\:\:{b}=\mathrm{4}.\mathrm{62}\:\:{c}=\mathrm{3}.\mathrm{38} \\ $$$$\:\:\:\:\:\:\alpha=\mathrm{95}.\mathrm{93}°\:\:\beta=\mathrm{50}°\:\:\gamma=\mathrm{34}.\mathrm{07}° \\ $$$$ \\ $$$$\mathrm{2}.\:{b}=\mathrm{6}\:\:\beta=\mathrm{50}°\:{b}+{c}=\mathrm{8}\:\Rightarrow\:{c}=\mathrm{2} \\ $$$$\alpha+\gamma=\mathrm{130}°\:\Rightarrow\:\gamma=\mathrm{130}°−\alpha \\ $$$${a}^{\mathrm{2}} =\mathrm{36}+\mathrm{4}−\mathrm{24cos}\:\alpha \\ $$$$\mathrm{36}={a}^{\mathrm{2}} +\mathrm{4}−\mathrm{4}{a}\mathrm{cos}\:\mathrm{50}° \\ $$$$\mathrm{4}={a}^{\mathrm{2}} +\mathrm{36}−\mathrm{12}{a}\mathrm{cos}\:\left(\mathrm{130}°−\alpha\right) \\ $$$$ \\ $$$$\Rightarrow\:{a}=\mathrm{7}.\mathrm{09}\:\:{b}=\mathrm{6}\:\:{c}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\alpha=\mathrm{115}.\mathrm{21}°\:\:\beta=\mathrm{50}°\:\:\gamma=\mathrm{14}.\mathrm{79}° \\ $$$$ \\ $$$$\mathrm{3}.\:{c}=\mathrm{6}\:\beta=\mathrm{50}°\:\:{b}+{c}=\mathrm{8}\:\Rightarrow\:{b}=\mathrm{2} \\ $$$$\alpha+\gamma=\mathrm{130}°\:\Rightarrow\:\gamma=\mathrm{130}°−\alpha \\ $$$${a}^{\mathrm{2}} =\mathrm{4}+\mathrm{36}−\mathrm{24cos}\:\alpha \\ $$$$\mathrm{4}={a}^{\mathrm{2}} +\mathrm{36}−\mathrm{12}{a}\mathrm{cos}\:\mathrm{50}° \\ $$$$\mathrm{36}={a}^{\mathrm{2}} +\mathrm{4}−\mathrm{4}{a}\mathrm{cos}\:\left(\mathrm{130}°−\alpha\right) \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{real}\:\mathrm{solution} \\ $$

Answered by MrW3 last updated on 05/Sep/18

Commented by MrW3 last updated on 05/Sep/18

This is how to draw the searched triangle.  1. draw line BC=6  2. draw line BD=8 with ∠B=50°  3. connect D with C  4. draw perpendicular bisector of DC which  intersects with BD at A  5. connect A with C  ΔABC is the searched triangle.  proof:  AC=AD  AB+AC=AB+AD=BD=8

$${This}\:{is}\:{how}\:{to}\:{draw}\:{the}\:{searched}\:{triangle}. \\ $$$$\mathrm{1}.\:{draw}\:{line}\:{BC}=\mathrm{6} \\ $$$$\mathrm{2}.\:{draw}\:{line}\:{BD}=\mathrm{8}\:{with}\:\angle{B}=\mathrm{50}° \\ $$$$\mathrm{3}.\:{connect}\:{D}\:{with}\:{C} \\ $$$$\mathrm{4}.\:{draw}\:{perpendicular}\:{bisector}\:{of}\:{DC}\:{which} \\ $$$${intersects}\:{with}\:{BD}\:{at}\:{A} \\ $$$$\mathrm{5}.\:{connect}\:{A}\:{with}\:{C} \\ $$$$\Delta{ABC}\:{is}\:{the}\:{searched}\:{triangle}. \\ $$$${proof}: \\ $$$${AC}={AD} \\ $$$${AB}+{AC}={AB}+{AD}={BD}=\mathrm{8} \\ $$

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