A) \[\text{7}0{}^\circ \]
B) \[\text{8}0{}^\circ \]
C) \[\text{9}0{}^\circ \]
D) \[\text{1}00{}^\circ \]
Correct Answer: D
Solution :
\[\sqrt{{{15}^{2}}+{{8}^{2}}}=\sqrt{289}=17\,m\] \['x'\] \[{{x}^{2}}={{34}^{2}}-{{16}^{2}}\] \[\Rightarrow \] \[x=\sqrt{{{34}^{2}}-{{16}^{2}}}=30\,m\] \[\Delta \text{ABC}\,\,\text{ }\!\!\tilde{\ }\!\!\text{ }\,\,\Delta \text{DEF}\] \[\frac{area\,of\,\Delta ABC}{area\,of\,\Delta DEF}={{\left( \frac{BC}{DE} \right)}^{2}}\]
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