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hammer [34]
3 years ago
11

A tourist starts off from town A and travels for 50km on a bearing of N80°W to town B. At town B, he continues for another 40km

on a bearing of N20°E to C. How far is C from A?
Mathematics
1 answer:
bixtya [17]3 years ago
3 0

Answer:

Town C is approximately 58.356 kilometers from town A.

Step-by-step explanation:

According to the statement, we understand that tourist travels from town A to town B (50 kilometers) at a direction of 80º west of north and from town B to town C (40 kilometers) at a direction of 20º east of north. Vectorially speaking, the resultant from town A to town C is described by the following formula:

\vec r = (50\,km)\cdot (-\sin80^{\circ},\cos 80^{\circ})+(40\,km)\cdot (\sin 20^{\circ}, \cos 20^{\circ})

\vec r = (-35.560,46.270)\,[km]

The distance from town A to town C is the magnitude of vector reported above, which is now calculated by Pythagorean Theorem:

\|\vec r\| = \sqrt{(-35.560\,km)^{2}+(46.270\,km)^{2}}

\|\vec r\| \approx 58.356\,km

Town C is approximately 58.356 kilometers from town A.

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o-na [289]

The parallel sides AB, PQ, and CD, gives similar triangles, ∆ABD ~ ∆PQD and ∆CDB ~ ∆PQB, from which we have;

\frac{1}{x}  +  \frac{1}{y}=   \frac{1}{z}

<h3>Which method can be used to prove the given relation?</h3>

From the given information, we have;

  • ∆ABD ~ ∆PQD
  • ∆CDB ~ ∆PQB

According to the ratio of corresponding sides of similar triangles, we have;

\frac{x}{z}  =  \mathbf{\frac{BD}{QD} }

\frac{y}{z}  =  \frac{BD}{ BQ}

Which gives;

\mathbf{\frac{y}{z}}  =  \frac{BD }{ BD  - Q D}

\frac{z}{y}  =  \frac{BD - QD }{ BD }  =   1 - \frac{Q D }{ BD}

QD × x = BD × z

BD × z = (1 - QD/BD) × y = y - (QD × y/BD)

Therefore;

BD × z = y - (QD × y/BD)

BQ × y = y - (QD × y/BD)

BQ × y = y - (z × y/x) = y × (1 - z/x)

(1 - z/x) = BQ

BD × z = y × (1 - z/x)

BD = (y × (1 - z/x))/z

Therefore;

QD × x = y × (1 - z/x)

(BD-BQ) × x = y × (1 - z/x)

(BD-(1 - z/x)) × x = y × (1 - z/x)

BD = (y × (1 - z/x))/x + (1 - z/x)

BQ + QD = (1 - z/x) + (y × (1 - z/x))/x

BD = BQ + QD

(y × (1 - z/x))/x + (1 - z/x) = (y × (1 - z/x))/z

(1 - z/x)×(y/x + 1) =(1 - z/x) × y/z

Dividing both sides by (1 - z/x) gives;

y/x + 1 = y/z

Dividing all through by y gives;

(y/x + 1)/y = (y/z)/y

  • 1/x + 1/y = 1/z

Therefore;

\frac{1}{x}  +  \frac{1}{y}=   \frac{1}{z}

Learn more about characteristics similar triangles here:

brainly.com/question/1799826

#SPJ1

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2 years ago
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