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標題:

Trigonometric ratios

發問:

In the figure, L is a lighthouse. A ship sailed from point A to point D along the dire-ction of AD with a uniform speed of 20 km/h. ABCD is a straight line,LA=110 km, LD=60 km and LD is perpendicular with ADSuppose the ship started at noon from point A and reached point B at 2:00p.m.(a)Find a and b.(b)If the... 顯示更多 In the figure, L is a lighthouse. A ship sailed from point A to point D along the dire-ction of AD with a uniform speed of 20 km/h. ABCD is a straight line,LA=110 km, LD=60 km and LD is perpendicular with ADSuppose the ship started at noon from point A and reached point B at 2:00p.m. (a)Find a and b. (b)If the ship reached point C at 3:00p.m., find theoter and LC. Picture:http://postimg.org/image/sm10wrodt/ NEED STEPS ,PLZ!!!

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In the figure, L is a lighthouse. A ship sailed from point A to point D along the dire-ction of AD with a uniform speed of 20 km/h. ABCD is a straight line,LA=110 km, LD=60 km and LD is perpendicular with AD. Suppose the ship started at noon from point A and reached point B at 2:00p.m. (a) Find a and b. sin a = 60/110 //a = 33.1°// (corr. to 3 s.f.) Length of AD = √(1102-602) (Pyth. theorem) = 10√85km Length of AB = Time × Speed = 2 × 20 = 40km Length of BD = AD - AB = (10√85-40)km tan b = 60/(10√85-40) //b = 49.0°//(corr. to 3 s.f.) (b) If the ship reached point C at 3:00p.m., find θ and LC. Length of AC = 3×20 = 60km Length of CD = AD - AC = (10√85-60)km Length of LC =√[(10√85-60)2+602] (Pyth. theorem) //= 68.1km// (corr. to 3 s.f.) cosθ = LD/LC =60/√[(10√85-60)2+602] //θ = 28.2°//(corr. to 3 s.f.)

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