# Solved Tests for Trigonometry

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The contents of this online page:

• - the problems 54 – 56 are presented with the examples of solutions and answers for the lesson "Trigonometric functions";
• - in the tests under numbers 57 – 63 of the workbook, it is considered how to solve trigonometry for the grade 9 if it is required to determine the sine, cosine, tangent, using the basic reduction formulas;
• - the problem of how to calculate the coordinates of a point having used the cosine and the sine of an angle, are explained in the test works for trigonometry 64 – 68;
• - the theme "formula of the area of a triangle, a parallelogram" is presented in the problems 69 – 71 of the textbook;
• - the law of sines is considered in the exercises for the geometry under numbers 72 – 74;
• - the law of cosines for a triangle is considered in the example 75.

## Trigonometric functions

The trigonometric ratios sine, cosine and tangent are defined as follows:

sine (sin) is opposite over hypotenuse,

cosine (cos) is adjacent over hypotenuse,

tangent (tg) is opposite over adjacent.

SOH, CAH, TOA.

Consider a right-angled triangle ΔOMD.

;

But OM=r=1, MD=y, OD=x

Then

Sin α = y for 0 ≤ y ≤ 1; 0 ≤ Sin α ≤ 1

Cos α=

Cos α = x for –1 ≤ x ≤ 1; –1 ≤ Cos α ≤ 1

Definition:

For any angles α (read: alpha): 0°≤ α≤180°

Sin α (the sine of α) is the ordinate of the point M,

Cos α (the cosine of α) is the abscissa of the point M.

Problem 54.

Given:

a circle (O;r),

coordinates of points A(1;0),

B(–1;0), M1(0;1); M2(;)

Find:

1) whether these points belong to the circle,

2) sine, cosine, tangent of angles

Solution:

1) Let us check the point A

x2 + y2 = r2

12 + 02 = 12

1=1 (true, this point belongs to the circle)

Let us check the point B

(–1)2 + 02 = 12

12 = 12

1=1 (true)

Check the point M1

02 + 12 = 12

1=1 (true)

Check the point M2

=(1)2

=1

=1

= 1

1=1 (true)

2) SinAOM1=1, CosAOM1=0

tgAOM1(not determined)

SinAOM2= , CosAOM2= ,

tgAOM2= = =

SinAOB=0, CosAOB=–1

tgAOM=== 0

0 90 180
sine Sin 0° = 0 sine Sin 90° = 1 sine Sin 180° = 0
cosine Cos 0° = 1 cosine Cos 90° = 0 cosine Cos 180° = –1
tangent tg 0° = = 0 tangent tg 90° ≠ (not determined) tangent tg 180° = 0

tg α =

The basic trigonometric identity (0°≤ α ≤180°)

The arc ACB is a part of the circle given by equation

x2 + y2 = 1, where

x = Cos α; y = Sin α

Then Cos2α + Sin2α = 1

***

Problem 55.

Given:

The cosine of the angle Cos α =

Calculate: the sine of the alpha angle Sin α = ?

Solution:

Cos2α + Sin2α = 1

Sin2α = 1 – Cos2α

Sin α =

Using the conditions 0°≤ α≤180°, 0 ≤ Sin α ≤ 1, we get

Sin α = ==

***

Problem 56.

Given:

The sine of the angle: Sin α =

Find: the cosine of the alpha angle: Cos α = ?

Solution:

Cos2α + Sin2α = 1

Cos2α = 1 – Sin2α

Cos α =

 Cos α = Cos α = – Cos α = Cos α = –

***

## The reduction formulas

Table. Trigonometric formulas of reduction. Sine, cosine in the trigonometry

 Sin (90° – α) = Cos α Cos (90° – α) = Sin α (0°≤ α≤90°) Sin (180° – α) = Sin α Cos (180° – α) = – Cos α (0°≤ α≤180°)

Problem 57.

Given:

The cosine of the angle Cos α =

Find: the sine of the alpha angle Sin α = ?

Solution:

Cos2α + Sin2α = 1

Sin2α = 1 – Cos2α

Sin α =

Using the condition 0 ≤ Sin α ≤ 1, we obtain

Sin α = ==

***

Problem 58.

Given:

The cosine of the angle Cos α =

Find: the sine of the alpha angle Sin α = ?

Solution:

Cos2α + Sin2α = 1

Sinα =

Sin α = ==

Using the condition 0 ≤ Sin α ≤ 1, we obtain

Sin α ==

***

Problem 59.

Given:

The sine of the angle Sin α =

Find: the cosine of the alpha angle Cos α = ?

Solution:

Cos2α + Sin2α = 1

Cos2α = 1 – Sin2α

Using the condition -1 ≤ Cos α ≤ 1, we obtain

Cos α = =

Cos α =

***

Problem 60.

Given:

The sine of the angle Sin α =0

Find: the cosine of the alpha angle Cos α = ?

Solution:

Cos2α + Sin2α = 1

Cos2α = 1 – Sin2α

Using the condition -1 ≤ Cos α ≤ 1, we obtain

Cos α = = ±1

***

Problem 61.

Given:

The cosine of the angle Cos α = 1

Find: tangent of the alpha angle tg α = ?

Solution:

tg α =

Cos2α + Sin2α = 1

Sin α = ±

Using the condition 0 ≤ Sin α ≤ 1, we obtain

Sin α = 0

Using the formula for the tangent, we obtain

tg α = tg α = 0

***

Problem 62.

Given:

The cosine of the angle Cos α =

Find: the tangent of the alpha angle tg α = ?

Solution:

tg α =

Cos2α + Sin2α = 1

Sin α = ±= ±

Using the condition 0 ≤ Sin α ≤ 1, we obtain

Sin α =

tg α == ==

***

Problem 63.

Given:

the angles 120°, 135°, 150°

Find: sine, cosine, tangent of the angles

Sin α =? Cos α = ? tg α = ?

Solution:

Sin 120° = Sin (180° - 60°) = Sin 60° =

Cos 120° = Cos (180° - 60°) = - Cos 60° = -

tg 120° = = • (-2) = -

Sin 135° = Sin (180° - 45°) = Sin 45° =

Cos 135° = Cos (180° - 45°) = - Cos 45° = -

tg 135° = = = -1

Sin 150° = Sin (180° - 30°) = Sin 30° =

Cos 150° = Cos (180° - 30°) = - Cos 30° = -

tg 150° = = = -1 • = -

***

## Coordinates of the point having used the cosine and the sine of the angle

Problem 64.

Given: a circle graph

The point A(x;y), y>0

A(x;y) does not lie on the circle (0;r)

The angle α

Prove:

x= OA • Cos α

y= OA • Sin α

Proof:

The point M lies on the circle of center at the origin and on the line OA, i.e.

M = the circle(0;r=1) ∩ OA or the circle(0;r=1) ∩ OA=M

The coordinates of the point M(x=Cos α; y=Sin α)

The coordinates of the vector {Cos α; Sin α}, since OM - is the radius vector of the point M.

Using the formula to calculate the length of the vector having used its coordinates, we get

====1

Then =OA•

{OA•Cos α; OA• Sin α} (*)

Since – is the radius vector of the point A, then

{x;y} (**)

Using conditions (*) and (**), we obtain

the formula is how to find the coordinates of a point having used the cosine and the sine of the angle:

x= OA • Cos α;

y= OA • Sin α

***

Problem 65.

Given:

OA=3

α= 45°

Find: the coordinates of the point A(x;y)

Solution:

Using the formulas for calculating the coordinate of a point having used the cosine and the sine of the angle, we get

x= OA • Cos α = 3 • Cos 45°= 3 • =

y= OA • Sin α = 3 • Sin 45°=

***

Problem 66.

Given:

the coordinates of the point A(2;2)

the circle of center at the origin (0;0)

Find: the angle α

Solution:

Since the given coordinates of the point A(2;2).

Using the formulas for calculating the coordinate of a point having used the cosine and the sine of the angle, we get

A (OA • Cos α; OA • Sin α);

Then OA • Cos α = 2

Using the formula to calculate the distance between two points , we get

OA = =

• Cos α = 2

Cos α = = α =45°

***

Problem 67.

Given: the length of the segment and the angle

 1) OA=1,5 α= 90° 2) OA=5 α= 150° 3) OA=2 α= 30°

Find: the coordinates of the point A(x;y)

Solution:

1) x= OA • Cos α = 1,5 • Cos 90°= 1,5 • 0 = 0

y= OA • Sin α = 1,5 • Sin 90°= 1,5 • 1 = 1,5

2) x= OA • Cos α = 5 • Cos (180° - 30°)= 5 • (- Cos 30°) = -

y= OA • Sin α = 5 • Sin (180 - 30°) = 5 • Sin 30° = 5 • == 2,5

3) x= OA • Cos α = 2 • Cos 30° = 2 • =

y= OA • Sin α = 2 • Sin 30° = 2 • = 1

Answer: 1) (0; 1,5) 2) (-; 2,5) 3) (;1)

***

Problem 68.

Given:

1) A(0;3)

2) B(–;1)

Find: the angle α

Solution:

1) Since the coordinates of the point A(0;3).

Using the formulas for calculating the coordinate of a point having used the cosine and the sine of the angle, we get

A (x=OA • Cos α; y=OA • Sin α);

Then x=OA • Cos α = 0

y= OA • Sin α = 3

Using the formula to calculate the distance between two points , we get

OA = = 3

3 • Cos α = 0

Cos α = 0

α = 90°

2) Since the coordinates of the point B(-;1).

Using the formulas for calculating the coordinate of a point having used the cosine and the sine of the angle, we get

B (x=OB • Cos α; y=OB • Sin α);

Then x = OB • Cos α = - (*)

Using the formula to calculate the distance between two points , we get

OB = == 2

Using the condition (*), we obtain

2 • Cos α = -

Cos α = -

α = 150°

Answer: 1) 90° ; 2) 150°

***

Theorem:

The area of the triangle is equal to half the product of the lengths of its two sides by the sine of the angle between them.

Given:

the triangle ΔABC

The sides of the triangle

BC=a, AC=b, AB=c

S - is the area of the triangle

Prove:

SΔABC=a • b • Sin C

Proof:

The area of the triangle is equal to half the product of the base of the triangle by its height. Therefore

SΔABC=a • h (*), where h – is the height of the triangle, a - is the base

But the height h = b • Sin C (**).

Using the condition (**) in the formula (*), then

The formula of the area of a triangle having used the sine, the base and the height

SΔABC=• a • b • Sin C

a = ; b =

***

Problem 69.

Given:

 1) AB = 6cm AC = 4 cm A = 60° 2) BC = 3 cm AB = 18cm B = 45° 3) AC = 14 cm CB = 7 cm C = 48°

Find: the area of the triangle SΔ

Solution:

Using the formula of the area of the triangle having used the sine, the base and the height, we get

SΔ=• a • b • Sin α; Sin 60° =

1) SΔ=• 6• 4 • = 6= 6= 12cm2

2) SΔ=• 3 • 18=== 27 cm2

3) SΔ=• 14 • 7 • Sin 48°, where Sin 48° ≈ 0,7

Then SΔ= 7 • 7 • 0,7 ≈ 34,3 cm2

Answer: 1) 12cm2 2) 27 cm2 3) 34,3 cm2

***

Problem 70.

Given:

The area of the triangle SΔABC= 60 cm2

the side AC = 15 cm

the angle A = 30°

Find: the length of the side AB

Solution:

SΔ= • AC • AB • Sin 30° = • AC • AB • = • AC • AB;

AB = = = 16 cm

***

Problem 71.

Theorem:

The area of the parallelogram is equal to the product of its two adjacent sides by the sine of the angle between them.

Given:

ABCD - is a parallelogram

AB = a

S - is the area of the parallelogram

Prove: S = a • b • SinA = a • b • Sin α

Proof:

Consider triangles ΔABD and ΔBDC. These triangles are equal by the third theorem.

Then the areas of the triangles are also equal SΔABD = SΔBDC

SΔBDC=• BC • CD • Sin α

Then the area of the parallelogram

SABCD= SΔABD + SΔBDC= 2 • SΔBDC

SABCD=2 • • a • b • Sin α = a • b • Sin α

***

## The law of sines

The law of sines for a triangle. Each side of an arbitrary triangle is proportional to the sine of the opposing angle.

Given:

The triangle ΔABC

AB=c, BC=a, AC=b

Prove:

Proof:

By the theorem about the area of a triangle, we obtain

SΔABC=c • b • Sin A (1)

SΔABC=a • b • Sin C (2)

SΔABC=c • a • Sin B (3)

It follows from (1) and (2) that

c • b • Sin A = a • b • Sin C

c • Sin A = a • Sin C

Then we obtain (*)

It follows from (2) and (3) that

a • b • Sin C =c • a • Sin B

b • Sin C = c • Sin B

Then we obtain (**)

From the equalities (*) and (**) we obtain

- is the formula of the sine of the triangle

***

Problem 72.

Given:

the triangle ΔABC

the side AC = 12 cm

the angle A = 75°

the angle С = 60°

How to find: the side AB of the triangle, and the area of the triangle SΔABC= ?

Solution: Using the law of sines, we obtain

(*)

B = 180° ─ (60° + 75°) = 45°

Substituting the known values of the variables in (*), we obtain

AB = = 6≈ 15 (cm)

Since the area of the triangle is equal to half the product of its two sides by the sine of the angle between them, we obtain the equality

SΔABC=• AB • AC • Sin A = • 15 • 12 • Sin 75° = 90 • 0,9659 ≈ 87 (cm2)

Answer: AB ≈ 15 cm, SΔABC ≈ 87 cm2.

***

Problem 73.

Given:

the triangle ΔABC

the side BC = cm

the angle A = 45°

the angle С = 30°

Find: the side AB of the triangle

Solution:

Using the law of sines, we obtain

AB = 1 cm.

***

Problem 74.

Given:

ABCD – is a rectangle

The diagonal AC = 10 cm

The angle BOC = 30°

Find: the area of a rectangle

SABCD=?

Solution:

Since ABCD - is a rectangle, we see that by the definition of a rectangle ABCD is a parallelogram. Consequently, it has the properties and attributes of a parallelogram.

AC=BD and BD=OD, AO=OC

Under the theorem about the parallelogram BO=OC; OD=AO.

Consider an isosceles triangle ΔBOC (BO=OC).

Since the angle BOC = 30°, we see that

OBC = BCO = (180° - 30°) : 2 = 75°, since the triangle Δ BOC - is an isosceles triangle.

Consider the triangle ΔBCD, where C = 90°.

The angle BDC = 180° - (75° + 90°) = 15°.

By the law of sines:

CD = 0,9659 • 10 = 9,6 (cm)

By the Pythagorean theorem

BC = = = = 2,8 (cm)

Then SABCD= BC • CD = 2,8 • 9,6 = 25 (cm2).

***

## The law of cosines

The law of cosines for a triangle. The square of either side of the triangle is equal to the sum of the squares of the other sides and minus the doubled product of these sides by the cosine of the angle between them.

Given:

the triangle ΔABC

AB=c, BC=a, AC=b

Prove:

a2 = b2 + c2 - 2bc • Cos A

Proof:

Consider a rectangular coordinate system, where

1) the point A - is the origin of the coordinate plane or A(0;0)

2) the side AB lies on the abscissa, i.e. the x-axis.

Then we get B(c;0), C (b • Cos A; b • Sin A)

Using the formula to calculate the distance between two points , we get

a2 = BC2 = (x2 - x1)2 + (y2 - y1)2 = (b • Cos A - c)2 + (b • Sin A - 0)2 =

= b2 • Cos2 A - 2bc • Cos A + c2 + b2 • Sin2 A = b2 (Cos2 A + Sin2 A) - 2bc • Cos A + c2 = b2 • 1 + c2 - 2bc • Cos A = b2 + c2 - 2bc • Cos A

Formulas of the law of cosines:

is the formula of the cosine of the triangle

***

Problem 75.

Given:

the triangle ΔABC

the height CH = hc

the height BH1 = hb

the angle A = α

Find: the area of the triangle

SΔABC = ?

Solution:

Let AB=c, AC=b

Then SΔABC =

Consider the triangle ΔABH1:

Sin α= c= (1)

Consider the triangle ΔAHC:

Sin α= b= (2)

Using equalities (1) and (2), we find the area of the triangle

SΔABC ==