Mathematics, 11.10.2019 12:30 shae5231

Ib precalculus: a ship is sailing north from point a towards point d. point c is 175 km north of a. point d is 60 km north of c. there is an island at e. the bearing of e from a is 055 degrees. the bearing of e from c is 134 degrees.

1. find the bearing of a from e

2. find ce

3. find de

Answers: 3

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Answer from: HyperZ

1. The bearing of A from E is 235 degree

2. The length of CE = 146 km

3. The length of DE = 193 km

Step-by-step explanation:

∵ The bearing of E from A is 055 degree

∴ The measure of the angle between North and segment AE

at E = 180 - 55 = 125°

∴ The bearing of A from E = 360 - 125 = 235 degree

1. The bearing of A from E is 235 degree

∵ The bearing of E from C is 134 degree

∴ The measure of the angle between North and segment CE

at E = 180 - 134 = 46°

∴ The measure of angle CEA = 125 - 46 = 79°

In triangle AEC:

∵ m∠A = 55° ⇒ given

∵ m∠AEC = 79° ⇒ proved

∵ AC = 175 km ⇒ given

* By using the sin Rule

∵ AC/sin(AEC) = CE/sin(A)

∴ 175/sin(79) = CE/sin(55) ⇒ by using cross multiplication

∴ CE = (175 × sin(55)) ÷ sin(79) = 146.04 ≅ 146 km

2. The length of CE = 146 km

In triangle DCE:

∵ m∠DCE = 134° ⇒ given

∵ DC = 60 km ⇒ given

∵ EC = 146 km ⇒ proved

* By using cos Rule:

∵ (DE)² = (CD)² + (CE)² - 2(CD)(CE) × cos(DCE)

∴ (DE)² = (60)² + (146.04)² - 2(60)(146.04) × cos(134) = 37099.04

∴ DE = 192.611 ≅ 193 km

3. The length of DE = 193 km

Answer from: Quest

$a_1 = 1$ , and

$a_{n+1} = a_n + (n + 1)$ for $n \ge 1$

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote $a_1 = 1$ , $a_2 = 3$ , $a_3 = 6$ , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from $a_1 = 1$ , to get to $a_2 = 3$ , we add $2(=1+1)$

from $a_2 = 3$ , to get to $a_3 = 6$ , we add $3 (=2+1)$

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

$a_1 = 1$ , and $a_{n+1} = a_n +(n + 1)$ for $n \ge 1$

Answer from: Quest

x = 2

step-by-step explanation:

if sr parallel to ba then the sides bc and ac are in proportion, that is

$\frac{sb}{cs}$ = $\frac{ra}{cr}$ ← substitute values

$\frac{4x+1}{2x}$ = $\frac{13.5}{6}$ ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = 2

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Ib precalculus: a ship is sailing north from point a towards point d. point c is 175 km north of a....