Trigonometry : corrected 3rd grade math exercises in PDF.

corrected by

The answer key to the math exercises in 3ème on trigonometry in the right triangle. Apply the sine, cosine and tangent formulas to calculate the length or measure of an angle.

Exercise 1:

we know that $\widehat{CAB}=50^{\circ}$ ; $\widehat{DBA}=15^{\circ}$ ; $\widehat{ACB}=90^{\circ}$ and $AB=40\,m$ .

Calculate the perimeter of triangle ABD. Round the result to the nearest decimeter.

In the right-angled triangle ABC :

$cos\widehat{A}=\frac{AC}{AB}$

$AC=40cos\widehat{50}$

$AC\simeq\,25,71\,m$

In the triangle ABD :

$\widehat{ADB}=180-50-15=180-65=115$

$\widehat{CDB}=180-115=65^{\circ}$

In the right-angled triangle ACB :

$sin\widehat{A}=\frac{BC}{AB}$

$BC=AB\times \,sinA$

$BC=40\times \,sin\,50^{\circ}$

$BC\simeq\,30,64\,m$

In the right triangle BCD :

$tan\widehat{BDC}=\frac{BC}{DC}$

$tan65=\frac{40\times \,sin50^{\circ}}{DC}$

$DC=\frac{40\times \,sin50^{\circ}}{tan65}$

$DC\simeq\,15,83\,m$

In addition

$DA=AC-DC=25,71-15,83=9,88\,m$

In the triangle BCD rectangular in C :

${\color{DarkRed}\,BD\simeq\,34,5\,m}$

The perimeter of triangle ABD is :

AD+DB+BA =9,88+34,5+40=84,38 m .

Conclusion: the perimeter is approximately 84.4 meters.

Exercise 2:

a. In the right-angled triangle DGE :

$sin\,\,\widehat{GED}=\frac{DG}{DE}$

$sin\,\,40=\frac{DG}{20}$

$DG=20sin\,\,40$

${\color{DarkRed}\,DG=12,9\,m}$

b. Represent the situation by the figure at 1/200 scale. (The data for the situation should be placed on the figure).

Exercise 3:

1.a. Using the calculator, calculate ( cos67°+sin67°)²+(cos67°-sin67°)²=2 (cos35°+sin35°)²+(cos35°-sin35°)²=2

b. what do we find?

The result is always equal to 2 .

2.prove that for any acute angle x :

$=2\times \,1\,\,(car\,cos^2x+sin^2x=1)$

Exercise 4:

Show that the triangle SON is a rectangle.

Calculation of the angle $\widehat{AOC}$ :

$cos\widehat{AOC}=\frac{3}{6}$

$\widehat{AOC}=cos^{-1}\frac{1}{2}$

$\widehat{AOC}=60^{\circ}$

The angles $\widehat{AOC}$ and $\widehat{EOS}$ are opposite by the vertex and therefore equal.

$\widehat{EOS}=60^{\circ}$

$\widehat{SON}=\widehat{EOS}+\widehat{NOE}=60^{\circ}+30^{\circ}=90^{\circ}$

Conclusion : the triangle NOS is a right-angled triangle in O .

Exercise 5:

is an angle such that $sinx=\frac{5}{8}$ .

$cos^2x=1-(\frac{5}{8})^2$

$cos^2x=1-\frac{25}{64}$

$cos^2x=\frac{64}{64}-\frac{25}{64}$

$cos^2x=\frac{39}{64}$

Now the cosine of an acute angle is positive:

$cosx=\sqrt{\frac{39}{64}}$

$cosx=\frac{\sqrt{39}}{8}$

$tanx=\frac{sinx}{cosx}$

$tanx=\frac{\frac{5}{8}}{\frac{\sqrt{39}}{8}}$

$tanx=\frac{5}{\sqrt{39}}$

$tanx=\frac{5\sqrt{39}}{39}$

Exercise 6:

1. Construct a triangle ABC at C such that AC = 5 cm and $\widehat{BAC}=40^{\circ}$ .

2. Calculate the length BC. (A value rounded to the nearest millimeter will be given).

According to the course sin $\widehat{BAC}$ = $\frac{BC}{AC}$ or sin 40° = BC/AC so BC = AC x sin 40° = 5 sin (40) $\approx$ 3,2cm

3.a) Where is the center O of the circumscribed circle of triangle ABC?

Since the triangle is right-angled, a property of the course says that the hypothenuse is a diameter of the circumscribed circle of the right-angled triangle(circumscribed meant that the circle passes through the three vertices of the triangle).

Now if [AC] is the diameter, then O is the middle of [AC].

b) Draw this circle.

4. Deduce the measure of the angle $\widehat{BOC}$ .

OB = OA so OAB is an isosceles triangle $\widehat{OBA}$ =40° implies that $\widehat{AOB}$ = 180°-(2×40°) since the sum of the angles of a triangle is always 180°. $\widehat{AOB}$ = 100° and since the angles $\widehat{AOB}$ and $\widehat{BOC}$ are supplementary (together they form a flat angle and their sum is 180°) we have $\widehat{BOC}$ = 180° -100° =80°.

Exercise 7:

What is the OH distance needed for the cathedral to appear fully in the lens?
I have the opposite side and the angle $\widehat{O}\,$ .
I am looking for the side adjacent to the angle $\widehat{O}\,$ .
Formula: tangent
$Tan\widehat{O}=\frac{AH}{OH}\,$

$Tan\,42=\frac{140}{OH}\,$

$OH=\frac{140}{Tan\,42}\,$

$OH=155,5\,m$

Conclusion:

The OH distance needed for the cathedral to appear fully in the lens must be greater than 155.5 meters.

Exercise 8:

a) The triangle SAH is right-angled at H.

So $\widehat{S}=90-45=45$ °

So the triangle SAH is a right triangle and isosceles in H.

BH=BA+AH=BA+HS=BA+x=40+x

b)AH=HS=x

c) In the triangle BSH rectangular in H.

$tan(\widehat{SBH})=\frac{SH}{BH}$

$tan(\widehat{SBH})=\frac{x}{40+x}$

$(40+x)tan(\widehat{SBH})=x$

${\color{DarkRed},(40+x)tan(\widehat{25})=x}$

d) $(40+x)tan(\widehat{25})=x$

$40tan(\widehat{25})+xtan(\widehat{25})=x$

$40tan(\widehat{25})+xtan(\widehat{25})=x,\\x(tan(\widehat{25})-1)=-40tan(\widehat{25}),\\x=\frac{-40tan(\widehat{25})}{tan(\widehat{25})-1}$

${\color{DarkRed},x\simeq,35\,\,m}$

The height of the keep is about 35 meters.

Exercise 9:

In the right block above, we give EH=69cm, EF=60cm and EA=51cm.
What is the measure of the angle AED? (round the result to the unit)

$tan\,\widehat{AED}=\frac{AD}{AE}$

$tan\,\widehat{AED}=\frac{69}{51}$

$\widehat{AED}=tan^{-1}(\frac{69}{51})$

$\widehat{AED}=54^{\circ}$

Exercise 10:

Help Lisa do this calculation with the help of the diagram below:

We have:

$tan\,40=\frac{BD}{BA}$ and $tan\,48=\frac{BD}{BC}$

using these two equalities

$BD=BA\times \,tan40=(BC+50)tan40$ and $BD=BC\times \,tan48$

Let’s determine BC :

$(BC+50)tan40=BC\times \,tan48$

$BC\times \,tan40+50tan40=BC\times \,tan48$

$BC\times \,tan40-BC\times \,tan48=-50tan40$

$BC=\frac{-50tan40}{tan40-tan48}$

Let’s determine BD :

$BD=BC\times \,tan48\\=\frac{-50tan40}{tan40-tan48}\times \,tan48\\\simeq\,171,62$

The archangel Saint Michael culminates at 171.62 meters.

Exercise 11:
1. Construct a full-size triangle ABC such that: AB = 7 cm; BC = 8 cm and AC = 5 cm.

2. [BC] being the side whose measure is the greatest, we should have if the triangle were rectangular in A :

BC² = AB² + AC²
or
8² $\not=$ 7² + 5²
So the triangle ABC is not rectangular.

2- Calculation of the angle $\hat{BAC}$ : let’s apply the formula

8² = 7² + 5² – 2*5*7 cos $\hat{BAC}$

64 = 49 + 25 – 70 cos $\hat{BAC}$

64 – 49 – 25 = -70 cos $\hat{BAC}$

-10 =-70 cos $\hat{BAC}$

cos $\hat{BAC}$ = $\frac{1}{7}$

Using the calculator we find :

$\hat{BAC}\,=\,81.79^\circ$

You can check this result using GEOGEBRA

Exercise 12:

1) For a 15% slope, what angle does the road make with the horizontal?

In this right-angled triangle, note $\widehat{A}$ the angle between the road and the horizontal.

We know the adjacent and opposite side of the angle $\widehat{A}$ , so the formula to use is the tangent.

$tan\widehat{A}=\frac{30}{100}=0,3$

$\widehat{A}=tan^{-1}(0,3)\simeq\,17^{\circ}$

Conclusion: the road makes an angle of about 17° with the horizontal.

2) A dangerous descent is considered as soon as the slope is higher than 10% on the road and higher than 4% on the highway.

From what angle between the road and the horizontal, is it considered dangerous to go downhill on a road?

$\widehat{A}=tan^{-1}(0,1)\simeq\,6^{\circ}$

$\widehat{B}=tan^{-1}(0,04)\simeq\,3^{\circ}$

Conclusion: a descent on a road is dangerous as soon as the angle is higher than 6° and higher than 3° for a highway.

3) Is it more dangerous to drive on a road with a 20% slope or to drive on a highway with a 20 degree angle to the horizontal? Justify

$\widehat{A}=tan^{-1}(0,2)\simeq\,3^{\circ}\simeq\,12^{\circ}$

Conclusion: On a highway the speed is much higher so it is more dangerous on a highway.

Exercise 13:
1°) Your triangle should look like this:

2°) To show that the triangle IJK is a right triangle,
we will use the reciprocal of the Pythagorean theorem.
The demonstration goes like this:

In the triangle IJK, we apply the reciprocal of the Pythagorean theorem, then we have :

JK² = 8² = 64 AND IJ² + IK ² = 4.8² + 6.4² = 23.04 + 40.96 = 64

Now JK² = IJ² + IK², so the triangle JIK is right-angled at I.

3°) We now want to know what is the measure of the angle $\widehat{IJK}\,$ .

Three possibilities of resolution, we use :

Exercise 14:

1°) We know that the wall (AB) and the floor are perpendicular.
We also know that the wall measures 3.05 m and that the scale [AC] measures 3.20m long.

So to know how far from the foot of the wall the ladder should be placed
so that its top is just at the level of the basket, we will use the Pythagorean theorem.
But first we convert: AB = 3.05m = 305cm and CA 3.20 m = 320 cm.

In the triangle ABC rectangle in B, we apply the Pythagorean theorem, we have :
CB²+AB²=CA²
CB²+305²=320²
CB²+93025=102400
CB²=102400-93025=9375
or CB>0 so $\fbox{CB=\sqrt{9375}\approx\,97\,cm}\,$

2°) The angle formed by the ladder and the ground is therefore the angle $\widehat{ACB}\,$ .

We have the three measures of the three sides of the triangle,
which gives us three possibilities.

Exercise 15:

1°) Your triangle should look like this one.

The angles $\widehat{AHC}\,$ and $\widehat{AHB}\,$ form two right angles because (AH) is the height of [BC] from vertex A.
Now the height is the straight line coming from a vertex and which is perpendicular to the opposite side.
We also know that BH = HC = BC/2 because in an isosceles triangle,
the height coming from the main vertex cuts its base in two equal parts because it is also a median.

2°) Calculation of $\widehat{B}\,$

It is known that the tangent of an angle is equal to the quotient of the opposite side of it by the adjacent side of it.
So:

$Tan\widehat{B}\,=\frac{AH}{BH}=\frac{7}{8/2}=\frac{7}{4}\,$

We deduce $\widehat{B}\,=\,Tan^{-1}(\frac{7}{4})\approx\,60\,$ degrees.

Exercise 16:
1°) A rectangle with its diagonal … no need for correction!!!

2°) Calculation of the measure of the angle $\widehat{ACD}\,$ :
We know the adjacent side and the opposite side of this angle,
which leads us to calculate the tangent of this angle.

$Tan\widehat{ACD}\,=\frac{AB}{BC}=\frac{7,2}{5,4}=\frac{4}{3}\,$

We deduce $\widehat{ACD}\,=\,Tan^{-1}(\frac{4}{3})\approx\,53\,$ degrees.

3°) Show that the angles $\widehat{ACD}\,$ and $\widehat{CAB}\,$ are equal. 1. method (the simplest) The lines (AB) and (DC) are parallel and the segment [AC] cuts $\widehat{BAD}\,$ and $\widehat{BCD}\,$ in two angles each.

We can therefore say that these two angles are internal alternates and therefore equal.

2. Method (for the addicts!!)
We calculate [AC] with Pythagoras:

In the right-angled triangle ACB (or ADC, they are the same), we apply the Pythagorean theorem:
AC²=AB²+BC²
AC²=7.2²+5.4²
AC²=51.84+29.16=81 or AC>0,therefore

$AC=\sqrt{81}=9\,$ cm .

We now have all the measurements of the sides of the rectangle.

So if the angles $\widehat{ACD}\,$ and $\widehat{CAB}\,$ were equal, the sine of one would be equal to the sine of the other and IDEM with the cosines.

Let’s check:

Indeed the angles $\widehat{ACD}\,$ and $\widehat{CAB}\,$ are equal.

Exercise 18:

Calculate, for each figure, the measure of the angle marked

(round the result to the nearest degree).

1. In the right triangle IAB, I know the side opposite and adjacent to the angle $\widehat{ABI}$ .

Formula : tangent.

$tan\widehat{ABI}=\frac{2,1}{2,8}$ so $\widehat{ABI}=tan^{-1}(\frac{2,1}{2,8})\simeq\,37^{\circ}$ .

2. In the right triangle DCL, I know the hypotenuse and opposite side of the angle $\widehat{DLC}$ .

Formula : sine.

$sin\,(\widehat{DLC})=sin(\frac{8}{9})$ so $\widehat{DLC}=sin^{-1}(\frac{8}{9})\simeq\,63^{\circ}$ .

3. In the right-angled triangle EFJ, I know the hypotenuse and opposite side of the angle $\widehat{JEF}$ .

Formula : sine.

$sin\,(\widehat{JEF})=\frac{2,7}{4,2}$ so $\widehat{JEF}=sin^{-1}(\frac{2,7}{4,2})\simeq\,40^{\circ}$ .

3. In the right triangle GHK, I know the side adjacent and opposite to the angle $\widehat{HKG}$ .

Formula : tangent.

$tan(\widehat{HKG})=\frac{4}{3}$ so $\widehat{HKG}=tan^{-1}(\frac{4}{3})\simeq\,53^{\circ}$

Exercise 20:

1. Calculate the measure of $\widehat{IGH}$ .

In the right triangle IGH, I know the side opposite to $\widehat{IGH}$ and the hypotenuse.

Formula : sine.

$sin(\widehat{IGH})=\frac{3}{6}$ so $\widehat{IGH}=sin^{-1}(\frac{3}{6})=30^{\circ}$ .

2. Deduce the measure of the angle $\widehat{EGF}$ .

The angles $\widehat{EGF}$ and $\widehat{IGH}$ are opposite by the vertex, so they have the same measure: $\widehat{EGF}=30$ °.

3. Calculate the lengths EF and FG rounded to the tenth.

In the triangle GEF rectangle in E.

$cos(\widehat{EGF})=\frac{EG}{FG}$ and $tan(\widehat{EGF})=\frac{EF}{EG}$

$cos(30^{\circ})=\frac{3}{FG}$ and $tan(30^{\circ})=\frac{EF}{3}$

$FG=\frac{3}{cos\,30^{\circ}}\simeq\,3,5$ cm.

$EF=3tan(30^{\circ})\simeq\,1,7$ cm

Exercise 21:

Calculate the length OM rounded to the millimeter.

Let’s calculate PM :

In the right triangle PAM, I know the opposite side and the angle $\widehat{APM}=47^{\circ}$

and I’m looking for the hypotenuse.

Formula: sine

$sin(\widehat{APM})=\frac{4,6}{PM}$

$sin(47^{\circ})=\frac{4,6}{PM}$ so

$PM=\frac{4,6}{sin(47^{\circ})}\simeq\,6,29\,cm$

Let’s calculate OM :

In the right triangle POM, I know the hypotenuse and the angle $\widehat{PMO}=23^{\circ}$

and I look for the side adjacent to the angle $\widehat{PMO}$ .

Formula: cosine.

$cos(\widehat{PMO})=\frac{OM}{PM}$

$cos(23^{\circ})=\frac{OM}{6,29}$

$OM\simeq\,6,29\times \,cos(23^{\circ})\simeq\,5,8\,cm$

Exercise 22:

We give BD = 4 cm , BA = 6 cm and $\widehat{DBC}=60^{\circ}$ .

1. Show that BC= 8 cm.

In the triangle DCB rectangle,

$cos60^{\circ}=\frac{DB}{BC}$

$cos60^{\circ}=\frac{4}{BC}$

$BC=\frac{4}{cos60^{\circ}}$

${\color{DarkRed}\,BC=8\,cm}$

2. Calculate CD.give the value rounded to the tenth.

$tan60^{\circ}=\frac{CD}{DB}$

$tan60^{\circ}=\frac{CD}{4}$

$CD=4tan60^{\circ}$

${\color{DarkRed}\,CD\simeq\,6,9\,cm}$

3. Calculate AC.

In the triangle ABC rectangle in B according to the direct part of the Pythagorean theorem :

$AC=\sqrt{100}$

${\color{DarkRed}\,AC=10\,cm}$

4. What is the value of $tan\widehat{BAC}$ ?

$tan\widehat{BAC}=\frac{BC}{BA}$

$tan\widehat{BAC}=\frac{8:2}{6:2}$

${\color{DarkRed}\,tan\widehat{BAC}=\frac{4}{3}}$

5. Deduce the value, rounded to the degree, of $\widehat{BAC}$ .

$\widehat{BAC}=tan^{-1}(\frac{4}{3})$

${\color{DarkRed}\,\widehat{BAC}=53^{\circ}}$

The answer key to the math exercises on trigonometry in the right triangle in 3rd grade.

After having consulted the answers to these exercises on trigonometry in 3rd grade, you can return to the exercises in 3rd grade

The exercises in the third grade.

Cette publication est également disponible en : Français (French) Español (Spanish) العربية (Arabic)

Download and print this document in PDF for free

You have the possibility to download then print this document for free «trigonometry : corrected 3rd grade math exercises in PDF.» in PDF format.

Other documents in the category corrected by

Download our free apps with all corrected lessons and exercises.

Other forms similar to trigonometry : corrected 3rd grade math exercises in PDF..

Statistics: answer key to 4th grade math exercises in PDF.
97
The answer key to the 4th grade math exercises on statistics and average in 4th grade. Know how to determine the median of a population and the character. Work on the classes of a continuous quantitative character. Exercise 1: 1) The entry [15;20[ refers to employees with 15 to 20…
Scratch: programming exercises in 5th grade.
97
Exercises with Scratch in order to work on the algorithm and programming part for the students of fifth grade in cycle 4. Assimilation of the different commands and bricks and understanding of algorithms. Exercise 1: Where is the cat when you click on the block? I click on but the…
Scratch : correction of the exercises in 3rd grade of programming.
96
The answers to the exercises on scratch in cycle 4 in the 3rd grade. Know how to create a program and set up an algorithm to answer a given problem. Exercise 1 Associate each program with the corresponding output. Program 1: plot 2. Program 2: plot 3. Program 3: plot…

Les dernières fiches mises à jour.

Voici les dernières ressources similaires à trigonometry : corrected 3rd grade math exercises in PDF. mis à jour sur Mathovore (des cours, exercices, des contrôles et autres), rédigées par notre équipe d'enseignants.

On Mathovore, there is 13 624 755 math lessons and exercises downloaded in PDF.