Question

2.3.19 Question Help A boat leaves the entrance to a harbor and travels 200 miles on a bearing of N 51° E. How many miles north and how many miles east from the harbor has the boat traveled? The boat has traveled miles north. (Round to the nearest tenth of a mile.)

Answer 1

From Given data The boat travel in North = 200xC0959 = 125.86 mi The boat travel in East = 200x sinsi = 155.43 mi 0 statue cet BC= x CD=l. 2 4350 120 > Faon de BCD. tan asso' = 1 - l= xx tan4350 - l = 0.96* & Fran de ACD. 120+* tan at = I. { substitute 4]. -> (120+x] tan zi = 0.964 =) .. . 52 61.14 +0.51% = 0.96%. -) 0.45% = 61.14 -) = 136.98m.. ::(.) = 0.96*= 0.96x 136.98 = 131.5 m. Height of statch or statue CD (1) = 131.5m.

= From Given data In shombus are equal all sides 12 m Fron de AEC casof = EC = 230 AC cas 67 = 6 - 180-90-23 = 60 6 II (a) 67 - 2 = 15.36 wat • c. E om The lengths of all sides of rhombus (x) = 15.36 m. -1002 = Given that Fron de BCE stan&u= BC =) EC = tandix 160 = 23.06 ft Pon De ECD 160.23KE-5647 TA tan 67.1 = C = CD = tan6#7 x 23:06 CE - 100 = 56:23 ft).

to height of airplane from top of tower cD = 56.234t height of airplane from bottan of tower BD = BD = BC + CD = 1604 56.23 = 216.23 ft. =) Given that rzo 20=60 22cmo zzem A 22cm BS 180-90-60 = 30. & sin so = y =) Y = 22x sin so =) y = 11cm. D C - cos sé = => x = 22x cay 38 => > = 19.05 cun. Area of De ADC - xy = 4 x 19.05 x 1 = 104.78 cm. . Ara od Ole ABC = = 2x alla of OADC 2 x 104.78 >A = 209.56 cm?