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Question Number 10690 by Saham last updated on 22/Feb/17

Answered by sandy_suhendra last updated on 23/Feb/17

Commented by sandy_suhendra last updated on 23/Feb/17

ΔTCB⇒tan 30°=(x/(TC)) TC=(x/(tan 30°))=1.732x ΔTCA⇒tan 60°=((20+x)/(TC)) TC=((20+x)/(tan 60°))=11.547+0.577x 1.732x=11.547+0.577x 1.155x=11.547 x=9.99 m a) AC=20+x=20+9.99=29.99 m ΔTCA⇒sin 60°=((AC)/(TA)) TA=((29.99)/(sin 60°))=34.63 m b) TF=AC=29.99 m

$$\Delta\mathrm{TCB}\Rightarrow\mathrm{tan}\:\mathrm{30}°=\frac{\mathrm{x}}{\mathrm{TC}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{TC}=\frac{\mathrm{x}}{\mathrm{tan}\:\mathrm{30}°}=\mathrm{1}.\mathrm{732x}\:\: \\ $$$$\Delta\mathrm{TCA}\Rightarrow\mathrm{tan}\:\mathrm{60}°=\frac{\mathrm{20}+\mathrm{x}}{\mathrm{TC}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{TC}=\frac{\mathrm{20}+\mathrm{x}}{\mathrm{tan}\:\mathrm{60}°}=\mathrm{11}.\mathrm{547}+\mathrm{0}.\mathrm{577x}\:\:\: \\ $$$$\mathrm{1}.\mathrm{732x}=\mathrm{11}.\mathrm{547}+\mathrm{0}.\mathrm{577x} \\ $$$$\mathrm{1}.\mathrm{155x}=\mathrm{11}.\mathrm{547} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{9}.\mathrm{99}\:\mathrm{m} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{AC}=\mathrm{20}+\mathrm{x}=\mathrm{20}+\mathrm{9}.\mathrm{99}=\mathrm{29}.\mathrm{99}\:\mathrm{m} \\ $$$$\:\:\:\:\:\Delta\mathrm{TCA}\Rightarrow\mathrm{sin}\:\mathrm{60}°=\frac{\mathrm{AC}}{\mathrm{TA}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{TA}=\frac{\mathrm{29}.\mathrm{99}}{\mathrm{sin}\:\mathrm{60}°}=\mathrm{34}.\mathrm{63}\:\mathrm{m}\:\:\: \\ $$$$\left.\mathrm{b}\right)\:\mathrm{TF}=\mathrm{AC}=\mathrm{29}.\mathrm{99}\:\mathrm{m} \\ $$

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