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

March 16, 2026Computer Science

ScanSolve AI Answer

March 16, 2026

Answer

96.42 m

Step 1: Determine $\angle ACB$ in $\triangle ABC$ .

\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 $AC$ 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 \cdot \sin \angle BAC

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

d = 150 \cdot \sin 40^\circ~m

Step 4: Compute $\sin 40^\circ \approx 0.6428$ .

d \approx 150 \times 0.6428 = 96.42~m

96.42 m

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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.

Asked on March 16, 2026Computer 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 AnswerAnswered on March 16, 2026

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