Answer:
Let PT be tower From \[\Delta \text{ }PTA\] \[\tan \,\,60{}^\circ =\frac{PT}{TA}\Rightarrow TA=\frac{15}{\sqrt{3}}\] from \[\Delta \text{ }PTB\] \[\tan \,\,45{}^\circ =\frac{PT}{TB}\Rightarrow TB=PT=15\,\,m\] Distance between two points \[AB=TB-TA\] \[=15-\frac{15}{\sqrt{3}}=\frac{15\left( \sqrt{3}-1 \right)}{\sqrt{3}}m\]
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