Sins Law: application, example and exercises

protection click fraud

THE law of sins determines that in any triangle, the sine relation of an angle is always proportional to the measure of the side opposite that angle.

This theorem demonstrates that in the same triangle the ratio between the value of one side and the sine of its opposite angle will always be constant.

Thus, for a triangle ABC with sides a, b, c, the Law of Sins admits the following relations:

Representation of the Laws of Sins in the triangle

Example

For a better understanding, let's calculate the measure of sides AB and BC of this triangle, as a function of measure b of side AC.

By the law of sines, we can establish the following relationship:

Hence, AB = 0.816b and BC = 1.115b.

Note: The values of sines were consulted in table of trigonometric ratios. In it, we can find the values of the angles from 1º to 90º of each trigonometric function (sine, cosine and tangent).

The angles of 30º, 45º and 60º are most used in trigonometry calculations. Hence, they are called remarkable angles. Check out a table with the values below:

instagram story viewer

Trigonometric Relations	30°	45°	60°
Sine	1/2	√2/2	√3/2
cosine	√3/2	√2/2	1/2
Tangent	√3/3	1	√3

Application of the Law of Sins

We use the Sine Law in acute triangles, where the internal angles are less than 90º (acute); or in obtuse triangles, which have internal angles greater than 90º (obtuse). In these cases, you can also use the Cosine Law.

The main objective of using the Law of Sins or Cosines is to discover the measurements of the sides of a triangle and also its angles.

Representation of triangles according to their internal angles

And the Law of Sins in the Rectangle Triangle?

As mentioned above, the Law of Sins is used in both acute and obtuse triangles.

In the right triangles, formed by an internal angle of 90º (straight), we used the Pythagorean Theorem and the relations between its sides: opposite, adjacent side and hypotenuse.

Representation of the right triangle and its sides

This theorem has the following statement: "the sum of the squares of their legs corresponds to the square of their hypotenuse". Its formula is expressed:

H² = ca²+ co²

Thus, when we have a right triangle, the sine will be the ratio between the length of the opposite leg and the length of the hypotenuse:

It reads opposite on the hypotenuse.

The cosine corresponds to the proportion between the length of the adjacent leg and the length of the hypotenuse, represented by the expression:

It is read adjacent to the hypotenuse.

Entrance Exam Exercises

1.(UFPB) The city hall of a certain city will build, over a river that crosses that city, a bridge that must be straight and connect two points, A and B, located on the opposite banks of the river. To measure the distance between these points, a surveyor located a third point, C, 200 m away from point A and on the same bank of the river as point A. Using a theodolite (a precision instrument for measuring horizontal angles and vertical angles, often used in topographic work), the surveyor observed that the angles B C with superscript logical conjunction A space and space C A with superscript logical conjunction B measured, respectively, 30º and 105º, as illustrated in the following figure.

Based on this information, it is correct to state that the distance, in meters, from point A to point B is:

a right parenthesis space 200 square root of 2 end space of root b right parenthesis space 180 square root of 2 end space of root c parenthesis right space 150 square root of 2 space d right parenthesis space 100 square root of 2 space and right parenthesis space 50 square root of 2

See Answer

R e s p o st a space c o r r e t a colon space d right parenthesis space 100 square root of 2

objective: Determine the measure of AB.

Idea 1 - Law of Sins to determine AB

The figure forms triangle ABC, where the side AC measures 200 m and we have two determined angles.

being the angle B with superscript logical conjunction opposite the side AC of 200 m and the angle C opposite the side AB, we can determine AB through the sins law.

$numerator A B over denominator s and n space 30 degree sign end of fraction space equal to space numerator A C about denominator s and n space start style show B with logical conjunction superscript end style end of fraction$

THE sins law determines that the ratios between the measurements of the sides and the sines of the opposite angles, respective to these sides, are equal in the same triangle.

Idea 2 - determine the angle

The sum of the interior angles of a triangle is 180°, so we can determine angle B.

B + 105° + 30° = 180°
B = 180° - 105° - 30°
B = 45°

Replacing the value of B with superscript logical conjunction in the law of sines and making the calculations.

$numerator A B space over denominator s and n space 30 degree sign end of fraction space equal to numerator space A C over denominator space s and n space B end of fraction numerator A B space over denominator s and n space 30 degree sign end of fraction space equal to numerator space A C over denominator space s e n space 45 degree sign end of fraction numerator A B space over denominator start style show 1 half end of style end of fraction space equal to numerator space A C over denominator space start style show numerator square root of 2 over denominator 2 end of fraction end of style end of fraction 2 A B space equal to numerator 2 A C over denominator square root of 2 end of fraction A B space equal to numerator A C over denominator square root of 2 end of fraction$

Note that there is a square root in a denominator. Let's take this root by doing the rationalization, which is the multiplication of both the denominator and numerator of the fraction by the root itself.

$A B space equal to numerator A C over denominator square root of 2 end of fraction space equal to space numerator A C space. square root space of 2 over denominator square root of 2 space. square root space of 2 end of fraction space equal to numerator space A C space. space square root of 2 over denominator square root of 4 end of fraction space equal to numerator space A C space. square root space of 2 over denominator 2 end of fraction$

Replacing the AC value, we have:

$A B space equal to space numerator 200 space. space square root of 2 over denominator 2 end of fraction space equal to space 100 square root of 2$

Therefore, the distance between points A and B is 100 square root of 2 m space .

2. (Mackenzie – SP) Three islands A, B and C appear on a map in 1:10000 scale, as shown in the figure. Of the alternatives, the one that best approximates the distance between islands A and B is:

a) 2.3 km
b) 2.1 km
c) 1.9 km
d) 1.4 km
e) 1.7 km

See Answer

Correct answer: e) 1.7 km

Purpose: To determine the measure of segment AB.

Idea 1: Use the sine law to find the measure of AB

Law of Sins: The measurements of the sides of a triangle are proportional to the sines of their opposite angles.

$numerator 12 over denominator s and n space 30 end of fraction space equal to space numerator A B over denominator space s and n space start style show C with logical conjunction superscript end style end of space fraction$

Idea 2: determine the angle

The sum of the interior angles of a triangle equals 180º.

30 + 105 + C = 180
135 + C = 180
C = 180 - 135
C = 45

Idea 3: Apply the value of C in the law of sines

$numerator 12 over denominator s and n space 30 end of fraction space equal to space numerator A B over denominator space s and n space start style show 45 end of style end of fraction space 12 space. space s and n space 45 space equal to space A B space. space s and n space 30 12 space. numerator space square root of 2 over denominator 2 end of fraction space equal to space A B space. space 1 middle 6 square root of 2 space equal to numerator A B over denominator 2 end of fraction 12 square root of 2 space equal to space A B$

Idea 4: approximate the square root value and use the scale

Making square root of 4 approximately equal space 1 comma 4

12. 1,4 = 16,8

The scale says 1:10000, multiplying:

16,8. 10000 = 168 000 cm

Idea 5: moving from cm to km

168 000 cm / 100 000 = 1.68 km

Conclusion: As the calculated distance is 1.68 km, the closest alternative is the letter e.

Note: To go from cm to km, we divide by 100 000 because, on the following scale, from centimeters to km, we count 5 places to the left.

km -5- hm -4- dam -3- m -2- dm -1- cm mm

3. (Unifor-CE) It is known that in every triangle the measure of each side is directly proportional to the sine of the angle opposite to the side. Using this information, it is concluded that the measure of side AB of the triangle shown below is:

a right parenthesis space 12 square root of 6 space m b right parenthesis space 12 square root of 3 space m c right parenthesis space 8 square root of 6 m space d right parenthesis space 8 square root of 3 m space and right parenthesis space 4 square root of 6 m space

See Answer

R e s p o st a space c o r r e t a colon space and right parenthesis space 4 square root of 6 space m.

The statement provides the law of sines.

$numerator 12 over denominator s and n space 120 end of fraction space equal to space numerator A B over denominator s and n space 45 end of fraction$

From trigonometry, we have that: sin 120 = sin 60.

Replacing the values in the formula:

$numerator 12 over denominator s and n space 120 end of fraction space equal to space numerator A B over denominator s and n space 45 end of fraction numerator 12 over denominator start style show numerator square root of 3 over denominator 2 end of fraction end of style end of fraction space equal to numerator A B over denominator start style show numerator square root of 2 over denominator 2 end of fraction end of style end of fraction 12 space. numerator space square root of 2 over denominator 2 end of fraction space equal to space A B space. numerator space square root of 3 over denominator 2 end of fraction 12 square root of 2 space equal to space A B square root of 3 A B space equal to space 12 numerator square root of 2 over denominator square root of 3 end of fraction$

In order not to leave a root in the denominator, we use rationalization, multiplying the denominator and the numerator by the root of 3.

$A B space equal to 12 space numerator square root of 2 over denominator square root of 3 end of space fraction. numerator space square root of 3 over denominator square root of 3 end of fraction space equal to space 12 numerator square root of 6 over denominator square root of 9 end of fraction space equal to space 12 numerator square root of 3 over denominator 3 end of fraction space equal to space 4 square root of 3$