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

Answered by MJS last updated on 03/May/18

$$\mathrm{aerial}:$$$$\mathrm{bottom}C=\left(\begin{array}{c}0\\ 0\end{array}\right)$$$$\mathrm{top}B=\left(\begin{array}{c}0\\ a\end{array}\right)$$$$\mathrm{surveyor}:$$$$\mathrm{position}1A_1=\left(\begin{array}{c}{b}_1\\ 0\end{array}\right)$$$$\mathrm{position}2A_2=\left(\begin{array}{c}{b}_1\\ -30\end{array}\right)$$$$\mathrm{triangle}{A}_{1}BC$$$$a=BC;b_1={CA}_1;c_1=A_{1}B$$$${\alpha }_1=\mathrm{\angle }{CA}_{1}B=48°;{\beta }_1=\mathrm{\angle }{A}_{1}BC=42°;{\gamma }_1=\mathrm{\angle }{BCA}_1=90°$$$$a^2+b_1^2=c_1^2$$$$\mathrm{triangle}{A}_{2}BC$$$$a=BC;b_2={CA}_2;c_2=A_{2}B$$$${\alpha }_2=\mathrm{\angle }{CA}_{2}B=44°;{\beta }_2=\mathrm{\angle }{A}_{2}BC=46°;{\gamma }_2=\mathrm{\angle }{BCA}_2=90°$$$$a^2+b_2^2=c_2^2$$$$\mathrm{triangle}{A}_2{CA}_1$$$$b_1={CA}_1;30=A_1{A}_2;b_2=A_{2}C$$$$b_1^2+900=b_2^2$$$$$$$$\frac{a}{\sin 48°}=\frac{b_1}{\sin 42°}\Rightarrow b_1=a\frac{\sin 42°}{\sin 48°}=a\times \tan 42°$$$$\frac{a}{\sin 44°}=\frac{b_2}{\sin 46°}\Rightarrow b_2=a\frac{\sin 46°}{\sin 44°}=a\times \tan 46°$$$${\left(a\times \tan 42°\right)}^2+900={\left(a\times \tan 46°\right)}^2$$$$a=\sqrt{\frac{900}{{\tan }^{2}46°-{\tan }^{2}42°}}\approx 58.66m$$

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