Question

Find the height of station B.

Answer 1

Using trigonometric relations for station $B$:
\[\tan 30^\circ = \frac{\text{Height of } B}{\text{Distance from } O}.\]
Substitute:
\[\tan 30^\circ = \frac{h_B}{36} \implies \frac{1}{\sqrt{3}} = \frac{h_B}{36}.\]
Solve for $h_B$:
\[h_B = \frac{36}{\sqrt{3}} = 12\sqrt{3} \, \text{m}.\]
Correct Answer: $12\sqrt{3}$ m.

Answer 2

Using trigonometric relations for station $A$:
\[\tan 45^\circ = \frac{\text{Height of } A}{\text{Distance from } O}.\]
Substitute:
\[\tan 45^\circ = \frac{h_A}{36} \implies 1 = \frac{h_A}{36}.\]
Solve for $h_A$:
\[h_A = 36 \, \text{m}.\]
Correct Answer: $36 \, \text{m}$.

Answer 3

Using the Pythagorean theorem for wire $OA$:
\[OA^2 = h_A^2 + OC^2.\]
Substitute:
\[OA^2 = 36^2 + 36^2 = 2(36^2) = 2(1296) = 2592.\]
\[OA = \sqrt{2592} = 36\sqrt{2} \, \text{m}.\]
Correct Answer: $36\sqrt{2}$ m.

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(iii) (b)Solution:
Using the Pythagorean theorem for wire $OB$:
\[OB^2 = h_B^2 + OC^2.\]
Substitute:
\[OB^2 = (12\sqrt{3})^2 + 36^2 = 432 + 1296 = 1728.\]
\[OB = \sqrt{1728} = 24\sqrt{3} \, \text{m}.\]
Correct Answer: $24\sqrt{3}$ m.