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Question-42942

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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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