Trigonometry exercises in the right triangle commented

Trigonometry is an important theme in Mathematics that makes it possible to know sides and angles in a right triangle, through the sine, cosine and tangent, in addition to other trigonometric functions.

To improve your studies and expand your knowledge, follow the list of 8 exercises, plus 4 entrance exam questions, all solved step by step.

Exercise 1

Observing in the morning the shadow of a building on the ground, one person found that it measured 63 meters when the sun's rays made an angle of 30° with the surface. Based on this information, calculate the height of the building.

Correct answer: Approximately 36.37 m.

The building, the shadow and the sun's ray determine a right triangle. Using the 30° angle and the tangent, we can determine the height of the building.

tan g e n t e space equal to numerator space c a t e t o space o po s t o over denominator c a t e t space a d j a c e n t e end of fraction

Since the height of the building is h, we have:

tan space 30 degree sign space equal to space h over 63 space space h space equal to space 63 space multiplication sign space tan space 30 degree sign space space space h space equal to space 63 space multiplication sign space numerator square root of 3 about denominator 3 end of fraction h space equal to space 21 square root of 3 space m h space approximately equal space 36 comma 37 space m

Exercise 2

On a circumference with a diameter of 3, a segment AC, called a chord, forms a 90° angle with another chord CB of the same length. What is the measure of the strings?

Correct answer: The length of the rope is 2.12 cm.

As the segments AC and CB form an angle of 90° and are of the same length, the triangle formed is isosceles and the base angles are equal.

Since the sum of the internal angles of a triangle is equal to 180° and we already have an angle of 90°, there are another 90° left to be divided equally between the two base angles. Thus, the value of these is equal to 45º each.

Since the diameter is equal to 3 cm, the radius is 1.5 cm and we can use the cosine of 45° to determine the length of the string.

cos space 45 degree sign space equal to space numerator 1 comma 5 over denominator c o r d end of fraction c o r d a space equal to space numerator 1 comma 5 over denominator cos space 45 degree sign end of fraction c or d a space equal to space numerator 1 comma 5 over denominator start style show numerator square root of 2 over denominator 2 end of fraction end of style end of fraction c o r d a space equals space 1 comma 5 space multiplication sign space numerator 2 over denominator square root of 2 end of fraction c or d a approximately equal space 2 comma 12 space cm

Exercise 3

A cyclist participating in a championship approaches the finish line at the top of a slope. The total length of this last part of the test is 60 m and the angle formed between the ramp and the horizontal is 30°. Knowing this, calculate the vertical height the cyclist needs to climb.

Correct answer: The height will be 30 m.

Calling the height of h, we have:

s and n space 30th space equal to space numerator h space over denominator 60 end of fraction space h space equal to space 60 space sign of multiplication space s and n 30 degree sign space h space equal to space 60 space multiplication sign space 1 half h space equal to space 30 m space

Exercise 4

The following figure is formed by three triangles where the height h determines two right angles. The element values ​​are:

α = 30°
β = 60°
h = 21

Find the value of a+b.

Right answer:

28 square root of 3

We can determine the measurements of segments a and b using the tangents of the given angles.

Calculation of a:

tan space alpha space equal to space a over h space space a space equal to space h space multiplication sign space tan alpha space space space a space equal to space 21 space multiplication sign space numerator square root of 3 over denominator 3 end of fraction space equal to 7 square root of 3

Calculation of b:

tan space beta space equal to space numerator b space over denominator h space end of fraction b space equal to space h space sign of multiplication space tan space beta b space equal to space 21 space multiplication sign space square root of 3 b space equal to 21 root square of 3

Thus,

a space plus space b space equals space 28 square root of 3

Exercise 5

A plane took off from city A and flew 50 km in a straight line until it landed in city B. Afterwards, it flew another 40 km, this time heading towards city D. These two routes are at a 90° angle to each other. However, due to unfavorable weather conditions, the pilot received a communication from the control tower informing him that he could not land in city D and that he should return to city A.

In order to make the U-turn from point C, the pilot would have to make a turn of how many degrees to the right?

Consider:

sin 51° = 0.77
cos 51° = 0.63
tan 51° = 1.25

Correct answer: The pilot must make a turn of 129° to the right.

Analyzing the figure, we see that the path forms a right triangle.

Let's call the angle we're looking for W. Angles W and Z are supplementary, that is, they form a shallow angle of 180°.

Thus, W + Z = 180°.

W = 180 - Z (equation 1)

Our task now is to determine the Z angle and, for that, we are going to use its tangent.

tan space Z space equal to space 50 over 40 tan space Z space equal to space 1 comma 25

We must ask ourselves: What is the angle whose tangent is 1.25?

The problem gives us this data, tan 51° = 1.25.

This value can also be found in a trigonometric table or with a scientific calculator, using the function:

tan to the power of minus 1 end of the exponential

Substituting the value of Z in equation 1, we have:

W = 180° - 51° = 129°

Exercise 6

A ray of monochromatic light when passing from one medium to another, suffers a deviation towards it. This change in its propagation is related to the refraction indices of the media, as shown in the following relationship:

Snell's Law - Descartes

s and n space r space x space n with 2 subscript space equal to space s and n space i space x space n with 1 subscript

Where i and r are the angles of incidence and refraction and, n1 and n2, the refractive indices of means 1 and 2.

When hitting the surface of separation between air and glass, a ray of light changes its direction, as shown in the figure. What is the refractive index of glass?

Data: Air refraction index equal to 1.

Correct answer: The refractive index of the glass is equal to square root of 3 .

Replacing the values ​​we have:

s and n space 30 degree sign space multiplication sign space n with vi i d r the subscript end of subscript space equal to space space n with a r subscript end of subscript space sign of multiplication space s and n space 60 degree sign space n with vi i d r the subscript end of subscript space equal to numerator space n with a r space subscript end of subscript sign of multiplication space s e n space 60 degree sign over denominator s e n space 30 degree sign end of fraction n with v i d r the subscript end of subscript space equal to space numerator 1 space multiplication sign start style show numerator square root of 3 over denominator 2 end fraction end style over denominator start style show 1 middle end style end of fraction n with v i d r the subscript end of subscript space equal to numerator space square root of 3 over denominator 2 end of fraction space multiplication sign space 2 over 1 space equal to square root space of 3

Exercise 7

To drag a wooden log into his workshop, a locksmith tied a rope to the log and pulled it ten feet across a horizontal surface. A force of 40 N through the string made an angle of 45° with the direction of travel. Calculate the work of the applied force.

Correct answer: The work performed is approximately 84.85 J.

Work is a scalar quantity obtained by the product of force and displacement. If the force does not have the same direction as the displacement, we must decompose this force and consider only the component in this direction.

In this case, we must multiply the magnitude of the force by the cosine of the angle.

So we have:

T space equals F space. space d space. space cos space 45 degree sign T space equals space 40 space. space 3 space. space numerator square root of 2 over denominator 2 end of fraction T space equal to space 60 space. 2 T square root space approximately equal space 84 comma 85 J space

Exercise 8

Between two mountains, the residents of two villages had to travel a hard way up and down. To resolve the situation, it was decided that a cable-stayed bridge would be built between villages A and B.

It would be necessary to calculate the distance between the two villages by the straight line on which the bridge would be stretched. As residents already knew the height of the cities and the angles of climb, this distance could be calculated.

Based on the diagram below and knowing that the height of the cities was 100 m, calculate the length of the bridge.

Correct answer: The bridge should have a length of approximately 157.73 m.

The bridge length is the sum of the sides adjacent to the given angles. Calling the height of h, we have:

Calculation with the 45° angle

tan space 45 degree sign space equal to space numerator h over denominator c a t e t the space a d j a c e n t and end of the fraction c a t e t the space a d j a c e n t e space equal to space numerator h over denominator tan space 45 degree sign end of fraction c a t e t space a d j a c e n t e equal space a space numerator 100 over denominator start style show 1 end of style end of fraction c a t e t space a d j a c e n t e space equal to 100 space m

Calculation with an angle of 60°

tan space 60 degree sign space equal to space numerator h over denominator c a t e t the space a d j a c e n t e end of the fraction c a t e t the space a d j a c e n t e space equal to space numerator h over denominator tan space 60 degree sign end of fraction c a t e t space a d j a c e n t e space equal to space numerator 100 over denominator start style show square root of 3 end of style end of fraction c a t e t space a d j a c e n t e space approximately equal space 57 comma 73 m space

To determine the bridge length, we sum the obtained values.

c o m pr i m e n t space equals space 100 space plus space 57 comma 73 space approximately equal space 157 comma 73 space m

question 1

Cefet - SP

In the triangle ABC below, CF = 20 cm and BC = 60 cm. Mark the measurements of the AF and BE segments respectively.

a) 5, 15
b) 10, 20
c) 15, 25
d) 20, 10
e) 10, 5

Answer: b) 10, 20

To determine AF

We note that AC = AF + CF, so we have to:

AF = AC - CF (equation 1)

CF is given by the problem, being equal to 20 cm.

AC can be determined using 30° sine.

s and n space 30 degree sign space equal to space numerator A C over denominator B C end of fraction space A C space equal to space B C space multiplication sign space s and n space 30 degree sign space

BC is provided by the problem, being equal to 60 cm.

A C space equals space 60 space multiplication sign space 1 half equals space 30 space c m.

Substituting in equation 1, we have:

A F space equals space A C space minus space C F space space A F space equals space 30 space minus space 20 space equals space 10 space c m

To determine BE

First observation:

We verify that the figure inside the triangle is a rectangle, due to the right angles determined in the figure.

Therefore, their sides are parallel.

Second observation:

The BE segment forms a right-angled triangle with an angle of 30° where: the height is equal to AF, which we have just determined, and BE is the hypotenuse.

Making the calculation:

We use 30° sine to determine BE

s and n space 30 degree sign space equal to 10 numerator space over denominator B E end of fraction space B space E space equal to 10 numerator space over denominator s and n space 30 degree sign end of fraction space B E space equal to space numerator 10 over denominator start style show 1 middle end of style end fraction B E space equal to space 20 space c m

question 2

EPCAR-MG

An airplane takes off from point B under a constant inclination of 15° to the horizontal. 2 km from B is the vertical projection C of the highest point D of a 600 m high mountain range, as shown in the figure.

Data: cos 15° = 0.97; sin 15° = 0.26; tg 15° = 0.27

It is correct to say that:

a) The plane will not collide with the saw before reaching 540 m in height.
b) There will be a collision between the plane and the saw at a height of 540 m.
c) The plane will collide with the saw in D.
d) If the plane takes off 220 m before B, keeping the same inclination, there will be no collision of the plane with the saw.

Answer: b) There will be a collision between the plane and the saw at a height of 540 m.

First, it is necessary to use the same multiple of the length measurement unit. Therefore, we will go 2 km to 2000 m.

Following the same initial flight conditions, we can predict the height at which the plane will be in the vertical projection of point C.

Using the 15° tangent and defining the height as h, we have:

tan space 15 degree sign space equal to space numerator h space over denominator 2000 end of fraction space h space equal to space 2000 space multiplication sign space tan space 15th space space h space equal to space 2000 space multiplication sign space 0 comma 27 space space space h space equal to space 540 space m

question 3

ENEM 2018

To decorate a straight circular cylinder, a rectangular strip of transparent paper will be used, on which a diagonal that forms 30° with the lower edge is drawn in bold. The radius of the base of the cylinder measures 6/π cm, and when winding the strip, a line in the shape of a helix is ​​obtained, as shown in the figure.

The measurement value of the height of the cylinder, in centimeters, is:

a) 36√3
b) 24√3
c) 4√3
d) 36
e) 72

Answer: b) 24√3

Observing the figure we notice that 6 turns were made around the cylinder. As it is a straight cylinder, anywhere in its height we will have a circle as the base.

To calculate the measure of the base of the triangle.

The length of a circle can be obtained from the formula:

Where r is the radius e, equal to typographic 6 on straight pi ,we have:

2 space. straight space pi space. space 6 space over straight pi

How are 6 laps:

6 space. space 2 space. straight space pi space. space 6 over straight pi space equals space 72 space

We can use the 30° tan to calculate height.

tan space 30 degree sign space equal to space numerator a l t u r a space over denominator b a s and end of fraction space space a l t u r a space equal to space b a s and space multiplication sign space tan space 30 degree sign space a l t u r a space equal to space 72 space multiplication sign space numerator square root of 3 over denominator 3 end of fraction a l t u r a space equal to space 24 square root of 3

question 4

ENEM 2017

Rays of sunlight are reaching the surface of a lake at an X angle with its surface, as shown in the figure.

Under certain conditions, it can be assumed that the luminous intensity of these rays, on the lake surface, is given approximately by I(x) = k. sin (x), k being a constant, and assuming that X is between 0° and 90°.

When x = 30º, the luminous intensity is reduced to what percentage of its maximum value?

A) 33%
B) 50%
C) 57%
D) 70%
E) 86%

Answer: B) 50%

Replacing the 30° sine value in the function, we obtain:

I left parenthesis x right parenthesis space equals space k space. s space and n space 30 degree sign I left parenthesis x right parenthesis space equal to space k space. 1 half space

Having reduced the value of k by half, the intensity is 50%.

Practice more exercises in:

Trigonometry Exercises

Expand your knowledge with:

Trigonometry in the right triangle

Metric Relations in the Rectangle Triangle

Trigonometry

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