Question-34230

Question Number 34230 by byaw last updated on 03/May/18

Answered by MJS last updated on 03/May/18

aerial: bottom C= ((0),(0) ) top B= ((0),(a) ) surveyor: position 1 A_1 = ((b_1 ),(0) ) position 2 A_2 = ((b_1 ),((−30)) ) triangle A_1 BC a=BC; b_1 =CA_1 ; c_1 =A_1 B α_1 =∠CA_1 B=48°; β_1 =∠A_1 BC=42°; γ_1 =∠BCA_1 =90° a^2 +b_1 ^2 =c_1 ^2 triangle A_2 BC a=BC; b_2 =CA_2 ; c_2 =A_2 B α_2 =∠CA_2 B=44°; β_2 =∠A_2 BC=46°; γ_2 =∠BCA_2 =90° a^2 +b_2 ^2 =c_2 ^2 triangle A_2 CA_1 b_1 =CA_1 ; 30=A_1 A_2 ; b_2 =A_2 C b_1 ^2 +900=b_2 ^2 (a/(sin 48°))=(b_1 /(sin 42°)) ⇒ b_1 =a((sin 42°)/(sin 48°))=a×tan 42° (a/(sin 44°))=(b_2 /(sin 46°)) ⇒ b_2 =a((sin 46°)/(sin 44°))=a×tan 46° (a×tan 42°)^2 +900=(a×tan 46°)^2 a=(√((900)/(tan^2 46°−tan^2 42°)))≈58.66m

$$\mathrm{aerial}: \\ $$$$\mathrm{bottom}\:{C}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{top}\:{B}=\begin{pmatrix}{\mathrm{0}}\\{{a}}\end{pmatrix} \\ $$$$\mathrm{surveyor}: \\ $$$$\mathrm{position}\:\mathrm{1}\:{A}_{\mathrm{1}} =\begin{pmatrix}{{b}_{\mathrm{1}} }\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{position}\:\mathrm{2}\:{A}_{\mathrm{2}} =\begin{pmatrix}{{b}_{\mathrm{1}} }\\{−\mathrm{30}}\end{pmatrix} \\ $$$$\mathrm{triangle}\:{A}_{\mathrm{1}} {BC} \\ $$$${a}={BC};\:{b}_{\mathrm{1}} ={CA}_{\mathrm{1}} ;\:{c}_{\mathrm{1}} ={A}_{\mathrm{1}} {B} \\ $$$$\alpha_{\mathrm{1}} =\angle{CA}_{\mathrm{1}} {B}=\mathrm{48}°;\:\beta_{\mathrm{1}} =\angle{A}_{\mathrm{1}} {BC}=\mathrm{42}°;\:\gamma_{\mathrm{1}} =\angle{BCA}_{\mathrm{1}} =\mathrm{90}° \\ $$$${a}^{\mathrm{2}} +{b}_{\mathrm{1}} ^{\mathrm{2}} ={c}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\mathrm{triangle}\:{A}_{\mathrm{2}} {BC} \\ $$$${a}={BC};\:{b}_{\mathrm{2}} ={CA}_{\mathrm{2}} ;\:{c}_{\mathrm{2}} ={A}_{\mathrm{2}} {B} \\ $$$$\alpha_{\mathrm{2}} =\angle{CA}_{\mathrm{2}} {B}=\mathrm{44}°;\:\beta_{\mathrm{2}} =\angle{A}_{\mathrm{2}} {BC}=\mathrm{46}°;\:\gamma_{\mathrm{2}} =\angle{BCA}_{\mathrm{2}} =\mathrm{90}° \\ $$$${a}^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} ={c}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{triangle}\:{A}_{\mathrm{2}} {CA}_{\mathrm{1}} \\ $$$${b}_{\mathrm{1}} ={CA}_{\mathrm{1}} ;\:\mathrm{30}={A}_{\mathrm{1}} {A}_{\mathrm{2}} ;\:{b}_{\mathrm{2}} ={A}_{\mathrm{2}} {C} \\ $$$${b}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{900}={b}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$ \\ $$$$\frac{{a}}{\mathrm{sin}\:\mathrm{48}°}=\frac{{b}_{\mathrm{1}} }{\mathrm{sin}\:\mathrm{42}°}\:\Rightarrow\:{b}_{\mathrm{1}} ={a}\frac{\mathrm{sin}\:\mathrm{42}°}{\mathrm{sin}\:\mathrm{48}°}={a}×\mathrm{tan}\:\mathrm{42}° \\ $$$$\frac{{a}}{\mathrm{sin}\:\mathrm{44}°}=\frac{{b}_{\mathrm{2}} }{\mathrm{sin}\:\mathrm{46}°}\:\Rightarrow\:{b}_{\mathrm{2}} ={a}\frac{\mathrm{sin}\:\mathrm{46}°}{\mathrm{sin}\:\mathrm{44}°}={a}×\mathrm{tan}\:\mathrm{46}° \\ $$$$\left({a}×\mathrm{tan}\:\mathrm{42}°\right)^{\mathrm{2}} +\mathrm{900}=\left({a}×\mathrm{tan}\:\mathrm{46}°\right)^{\mathrm{2}} \\ $$$${a}=\sqrt{\frac{\mathrm{900}}{\mathrm{tan}^{\mathrm{2}} \:\mathrm{46}°−\mathrm{tan}^{\mathrm{2}} \:\mathrm{42}°}}\approx\mathrm{58}.\mathrm{66}{m} \\ $$

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