A submarine travels 15 km due East from its base and then turns and travels due North for 6.8 km. How far away is the submarine from its base? Give your answer rounded to 1 DP.
Answers
♧QUESTION♧
A submarine travels 15 km due East from its base and then turns and travels due North for 6.8 km. How far away is the submarine from its base? Give your answer rounded to 1 DP.
☆ANSWER☆
{\bold{\therefore Distance=13.32\:km\:\:North-West}}∴Distance=13.32kmNorth−West
{\bold{\underline{\underline{Step-by-step\:explanation:}}}}Step−by−stepexplanation:
• In the given question information given about a submarine travels 10 km due North from its base and then turns and travels due West for 8.8 km.
• We have to find the distance from submarine to its base.
\begin{gathered}\underline \bold{Given : } \\ \implies AB = 10 \: km \\ \\ \implies BC = 8.8 \:km \\ \\ \underline \bold{to \: find : } \\ \implies AC = ?\end{gathered}Given:⟹AB=10km⟹BC=8.8kmtofind:⟹AC=?
• According to given question :
\begin{gathered}\bold{ \angle B = 90 \degree}\\ \\ \bold{Using \: Phythagoras \: theoram : } \\ \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \implies {AC}^{2} = {AB}^{2} + {BC}^{2} \\ \\ \implies {AC}^{2} = ( {10})^{2} + {(8.8)}^{2} \\ \\ \implies {AC}^{2} = 100 + 77.44 \\ \\ \implies {AC}^{2} = 177.44 \\ \\ \implies AC= \sqrt{177.44} \\ \\ \bold{\implies AC = 13.32 \: km} \\ \\ \bold{\therefore Submarine \: is \: 13.32 \: km \:\: North -West\: away \: from \: base}\end{gathered}∠B=90°UsingPhythagorastheoram:⟹h2=p2+b2⟹AC2=AB2+BC2⟹AC2=(10)2+(8.8)2⟹AC2=100+77.44⟹AC2=177.44⟹AC=177.44⟹AC=13.32km∴Submarineis13.32kmNorth−Westawayfrombase