Question Number 73901 by azizullah last updated on 16/Nov/19 | ||
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Answered by $@ty@m123 last updated on 16/Nov/19 | ||
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$$\mathrm{42}\:{m} \\ $$ | ||
Commented by azizullah last updated on 16/Nov/19 | ||
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$$\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{work}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{solution}}. \\ $$ | ||
Answered by $@ty@m123 last updated on 16/Nov/19 | ||
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$${Let}\:{AB}\:{be}\:{a}\:{tower}.\:\left({Draw}\:{it}.\right) \\ $$$${Let}\:{BC}=\mathrm{42}\:{m}\: \\ $$$${ATQ} \\ $$$$\angle{ACB}=\mathrm{45}^{\mathrm{o}} \\ $$$$\Rightarrow\:\angle{BAC}=\mathrm{45}^{\mathrm{o}} \\ $$$$\Rightarrow{AB}={BC} \\ $$$$\Rightarrow{AB}=\mathrm{42}\:{m} \\ $$ | ||
Commented by azizullah last updated on 17/Nov/19 | ||
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$$\:\:\:\:\:\mathrm{thanks}\:\mathrm{alot}\:\mathrm{sir}\:\mathrm{G} \\ $$ | ||