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Four ships A, B, C and D are at sea in the following relative positions : B is on the straight line segment AC, B is due North of D and D is due west of C. The distance between B and D is 2 km. ∠BDA = 40°, ∠ BCD = 25°. What is the distance between A and D ? [Take sin 25° = 0.423]

Four ships A, B, C and D are at sea in the following relative positions : B is on the straight line segment AC, B is due North of D and D is due west of C. The distance between B and D is 2 km. ∠BDA = 40°, ∠ BCD = 25°. What is the distance between A and D ? [Take sin 25° = 0.423]

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
7 years ago
Hello Student,
Please find the answer to your question
According to question figure is as follows
236-1694_12345.png
Here, ∠ BDC = 90°, BD = 2 km
∠BDA = 40° ⇒ ∠ADC = 130°
∴ ∠DAC = 180° - (25° + 130°) = 25°
From the figure, in ∆ ABD, using sine law
AD/sin 115° = BD/sin 25° ⇒
AD = 2 sin (90° + 25°)/sin 25° = 2 cos 25°/sin 25°
⇒ AD = 2 cot 25°
= 2 √1/sin2 25° - 1 = 2√1/(0.423)2 – 1 = 4.28 km

Thanks
Aditi Chauhan
askIITians Faculty

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