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

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 · sin ∠ BAC; Step 3: Substitute AB = 150 m and ∠ BAC = 40°. d = 150 · sin 40°~m Step 4: Compute sin 40° ≈ 0.6428. d ≈ 150 × 0.6428 = 96.42~m 96.42 m

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

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 · sin ∠ BAC; Step 3: Substitute AB = 150 m and ∠ BAC = 40°. d = 150 · sin 40°~m Step 4: Compute sin 40° ≈ 0.6428. d ≈ 150 × 0.6428 = 96.42~m 96.42 m

Solved: In the space below make a neat orthographic drawing…

Determine $\angle ACB$ in $\triangle ABC$ .
$\angle ACB = 180^\circ - \angle BAC - \angle ABC = 180^\circ - 40^\circ - 60^\circ = 80^\circ$
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 $AC$ .
$d = AB \cdot \sin \angle BAC$
Substitute $AB = 150$ m and $\angle BAC = 40^\circ$ .
$d = 150 \cdot \sin 40^\circ~\text{m}$
Compute $\sin 40^\circ \approx 0.6428$ .
$d \approx 150 \times 0.6428 = 96.42~\text{m}$
96.42 m

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

Answer

Step-by-step solution

Determine $\angle ACB$ in $\triangle ABC$ .

The banks are parallel straight lines, with line $AC$ along one bank and point $B$ on the other bank.

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

Compute $\sin 40^\circ \approx 0.6428$ .

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

Answer

Step-by-step solution

Determine ∠ACB\angle ACB∠ACB in △ABC\triangle ABC△ABC.

The banks are parallel straight lines, with line ACACAC along one bank and point BBB on the other bank.

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

Compute sin⁡40∘≈0.6428\sin 40^\circ \approx 0.6428sin40∘≈0.6428.

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Still stuck on this one?

More Computer Science Questions

Determine $\angle ACB$ in $\triangle ABC$ .

The banks are parallel straight lines, with line $AC$ along one bank and point $B$ on the other bank.

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

Compute $\sin 40^\circ \approx 0.6428$ .