In the space below make a neat orthographic drawing to a scale 2:1 of the given isometric view.

Computer Science
In the space below make a neat orthographic drawing to a scale 2:1 of the given isometric view.

ScanSolve AI Answer

Answer

96.42 m

Step 1: Determine ∠ACB\angle ACB in △ABC\triangle ABC.

∠ACB=180∘−∠BAC−∠ABC=180∘−40∘−60∘=80∘\angle ACB = 180^\circ - \angle BAC - \angle ABC = 180^\circ - 40^\circ - 60^\circ = 80^\circ

Step 2: The banks are parallel straight lines, with line ACAC along one bank and point BB on the other bank. The bridge length dd is the perpendicular distance from BB to line ACAC.

d=AB⋅sin⁡∠BACd = AB \cdot \sin \angle BAC

Step 3: Substitute AB=150AB = 150 m and ∠BAC=40∘\angle BAC = 40^\circ.

d=150⋅sin⁡40∘ md = 150 \cdot \sin 40^\circ~m

Step 4: Compute sin⁡40∘≈0.6428\sin 40^\circ \approx 0.6428.

d≈150×0.6428=96.42 md \approx 150 \times 0.6428 = 96.42~m

96.42 m

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Quick Answer

Determine ACB in ABC. ACB = 180^ - BAC - ABC = 180^ - 40^ - 60^ = 80^ Step 2: The banks are parallel straight lines, with line AC along one bank and point B on the other bank.

In the space below make a neat orthographic drawing to a scale 2:1 of the given isometric view.
Computer Science

This computer science problem involves algorithmic thinking and programming concepts. The solution below explains the approach, logic, and implementation step by step.

ScanSolve AI Answer
Step 1: Determine ACB in ABC. ACB = 180^ - BAC - ABC = 180^ - 40^ - 60^ = 80^ Step 2: The banks are parallel straight lines, with line AC along one bank and point B on the other bank. The bridge length d is the perpendicular distance from B to line AC. d = AB · BAC Step 3: Substitute AB = 150 m and BAC = 40^. d = 150 · 40^~m Step 4: Compute 40^ ≈ 0.6428. d ≈ 150 × 0.6428 = 96.42~m 96.42 m