θ = 3 \theta = 3^\circ

Geometry Level 2

Given that the radius of earth is 6428 km. A and B creates a 3 3^\circ angle in centre of earth.

Find the distance between A and B to 2 decimal places.(In km).


The answer is 336.57.

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1 solution

Wenjin C.
May 21, 2018

See Sine Rule

Sine Law (Trigonometry)

Note: The author of the problem didn’t make clear that points A and B are points on the sphere... but here I’ll present my solution to this problem with an added condition that A and B are points on the sphere... until this condition has been incorporated into it, this solution will not be appropriate with the given conditions stated in the problem.

Note : In this solution, L \angle L denotes the angle L and not the measurement of it, while m L m\angle L denotes the measurement of angle L. Also, in this solution, M N \overline{MN} may either refer to the line MN or refer to the line segment MN but not the measurement or length of it; on the other hand, M N MN refers to the length of segment MN.

Name the center point O .

According to the problem, the length of the radius of the given sphere (assuming earth is a perfect sphere for the sake of the problem) is 6428 km. The problem also states that points

A and B create a 3 ° angle in centre of earth.

m A O B = 3 ° \Rightarrow m\angle AOB = 3° .

By law of sines, the lengths of the sides and the measures of the angles in a triangle X Y Z XYZ can be put into this relationship:

Y Z s i n X = Z X s i n Y = X Y s i n Z \frac {YZ}{sinX} = \frac {ZX}{sinY} = \frac {XY}{sinZ} .

Connect the center O and points A and B together to form triangle AOB. (Definition of a triangle)

Assuming that A and B are points on the sphere, we have:

segments OA and OB are both radii (radiuses) of the earth (sphere) and therefore O A \overline{OA} is congruent to O B \overline{OB} . (Radius of a sphere is congruent to another radius of the same sphere.)

Thus, by definition of congruent segments, the length of O A \overline{OA} is equal to the length of O B \overline{OB} .

\Rightarrow O A = O B = 6428 k m OA=OB=6428 km . (From given that the radius measures 6428 km) Also, triangle O A B OAB is an isosceles triangle since O A \overline{OA} is congruent to O B \overline{OB} . (Definition of an isosceles triangle)

B A O \Rightarrow \angle BAO is congruent to O B A \angle OBA .

(Property of isosceles triangles: Given an isosceles triangle DEF, where segment DE is congruent to segment DF, angle E is also congruent to angle F. This property can be proved by the law of sines, too: D E s i n F = D F s i n E D E s i n F = D E s i n E \frac {DE}{sinF} = \frac {DF}{sinE} \Rightarrow \frac {DE}{sinF} = \frac {DE}{sinE} (by law of substitution) s i n E = s i n F \Rightarrow sinE = sinF (by Multiplication Property of Equality or MPE) \Rightarrow (since we are just dealing with triangles and all three angles of the polygon measure less than 180°) m E = m F E m\angle E = m\angle F \Rightarrow \angle E is congruent to F \angle F (by definition of congruent angles).)

m B A O = m O B A \Rightarrow m\angle BAO = m\angle OBA (Definition of congruent angles: If angle P is congruent to angle S, then their measurements are equal.)

The sum of the measurements of all angles in a triangle is 180°. In the formed triangle here named AOB, m A O B + m B A O + m O B A = 180 ° m\angle AOB + m\angle BAO + m\angle OBA = 180° .

Since m B A O = m O B A m\angle BAO = m\angle OBA and m A O B = 3 ° m\angle AOB = 3° , by substitution, we have:

3 ° + m O B A + m O B A = 180 ° 3° + m\angle OBA + m\angle OBA = 180° 3 ° + 2 m O B A = 180 ° \Rightarrow 3° + 2m\angle OBA = 180° .

By Addition Property of Equality (APE), add 3 ° -3° to both the LHS (left-hand side) and RHS (right-hand side) of the equation.

2 m O B A = 177 ° \Rightarrow 2m\angle OBA = 177° .

By MPE:

m O B A = 88.5 ° \Rightarrow m\angle OBA = 88.5° .

The distance from A to B is desired, therefore the value of A B AB is to be solved here. Applying sine law on triangle AOB, we have:

A B s i n ( m A O B ) = O A s i n ( m O B A ) \frac {AB}{sin(m\angle AOB)} = \frac {OA}{sin(m\angle OBA)} .

Applying substitution of known values,

A B s i n ( 3 ° ) = 6428 k m s i n ( 88.5 ° ) \Rightarrow \frac {AB}{sin(3°)} = \frac {6428 km}{sin(88.5°)} .

Applying MPE:

A B = 6428 s i n ( 3 ° ) s i n ( 88.5 ° ) k m \Rightarrow AB = \frac {6428sin(3°)}{sin(88.5°)} km .

A B \Rightarrow AB is around 336.5308474 k m \boxed{336.5308474 km} . (Approximate answer using calculator.)

Rounded to two decimal places, the answer is 336.53 k m \boxed{336.53 km} .

———

Another way, in a picture:

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