Geometry and Trigonometry Alıştırma Soruları — JAMB UTME

Soru 1Soru →

For the interval $0^\circ \le x \le 360^\circ$ , solve the trigonometric equation $2\cos^2 x - \cos x - 1 = 0$ . Which set contains all the solutions for $x$ ?

0^\circ, 120^\circ, 240^\circ, 360^\circ

60^\circ, 120^\circ, 240^\circ, 300^\circ

0^\circ, 120^\circ, 360^\circ

60^\circ, 300^\circ, 360^\circ

Cevabı ve açıklamayı göster

Cevap: $0^\circ, 120^\circ, 240^\circ, 360^\circ$

Cevap

The complete set of solutions is

0^\circ, 120^\circ, 240^\circ, 360^\circ

.

Factoring

2\cos^2 x - \cos x - 1 = 0

gives

(2\cos x + 1)(\cos x - 1) = 0

. Setting the first factor to zero yields

\cos x = -\frac{1}{2}

, which has solutions at

120^\circ

and

240^\circ

in the interval

[0^\circ, 360^\circ]

. Setting the second factor to zero gives

\cos x = 1

, which has solutions at

0^\circ

and

360^\circ

. Combining these yields the set

0^\circ, 120^\circ, 240^\circ, 360^\circ

.

Adım Adım Çözüm

1

Factor the quadratic trigonometric equation

(2\cos x + 1)(\cos x - 1) = 0

Treat

\cos x

as a single variable to simplify into standard quadratic factors.

2

Set each factor to zero to find values for

\cos x

\cos x = 1

or

\cos x = -\frac{1}{2}

Zero-product property requires at least one factor to be zero.

3

Solve for

x

in the interval

0^\circ \le x \le 360^\circ

For

\cos x = 1

:

x = 0^\circ, 360^\circ

. For

\cos x = -\frac{1}{2}

:

x = 180^\circ - 60^\circ = 120^\circ

(Quadrant II) and

x = 180^\circ + 60^\circ = 240^\circ

(Quadrant III).

Cosine is negative in Quadrants II and III, with a reference angle of

60^\circ

.

4

Combine all unique solutions in ascending order

x = 0^\circ, 120^\circ, 240^\circ, 360^\circ

Include all solutions within the given domain boundaries.

Anahtar Kavram

Solving Quadratic Trigonometric Equations

Soru 2Soru →

Find the smallest positive value of $\theta$ (in degrees) satisfying the trigonometric equation $\sin(3\theta) + \sqrt{3}\cos(3\theta) = \sqrt{2}$ .

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

The smallest positive value of

\theta

is

25^\circ

.

Dividing the equation by

2

reduces it to

\sin(3\theta + 60^\circ) = \frac{\sqrt{2}}{2}

. The principal acute angle is

45^\circ

. The second quadrant angle giving a positive sine is

180^\circ - 45^\circ = 135^\circ

. Setting

3\theta + 60^\circ = 135^\circ

gives

3\theta = 75^\circ

, which yields

\theta = 25^\circ

. This is smaller than any positive solution generated by the first quadrant branch.

Adım Adım Çözüm

1

Transform the left-hand side into a single harmonic function

R\sin(3\theta + \alpha)

.

Dividing the equation by

2

yields

\frac{1}{2}\sin(3\theta) + \frac{\sqrt{3}}{2}\cos(3\theta) = \frac{\sqrt{2}}{2}

, which simplifies to

\sin(3\theta + 60^\circ) = \frac{\sqrt{2}}{2}

.

The identity

\sin(A + B) = \sin A \cos B + \cos A \sin B

allows us to combine

\sin(3\theta)

and

\cos(3\theta)

using

\cos 60^\circ = \frac{1}{2}

and

\sin 60^\circ = \frac{\sqrt{3}}{2}

.

2

Find the quadrant solutions for the angle

3\theta + 60^\circ

.

First quadrant:

3\theta + 60^\circ = 45^\circ + 360^\circ k \implies 3\theta = -15^\circ + 360^\circ k

.
Second quadrant:

3\theta + 60^\circ = 135^\circ + 360^\circ k \implies 3\theta = 75^\circ + 360^\circ k

.

Since the sine of the angle is positive (

\frac{\sqrt{2}}{2}

), solutions exist in both the first (

45^\circ

) and second (

180^\circ - 45^\circ = 135^\circ

) quadrants.

3

Calculate the smallest positive angle

\theta

.

For

k = 0

in the second quadrant branch,

3\theta = 75^\circ \implies \theta = 25^\circ

. (The first quadrant branch yields

\theta = -5^\circ

for

k=0

and

\theta = 115^\circ

for

k=1

). Thus,

\theta = 25^\circ

is the smallest positive solution.

Comparing all non-negative resulting angles demonstrates that

25^\circ

is the smallest strictly positive solution.

Anahtar Kavram

Solving linear trigonometric equations of the form

A\sin x + B\cos x = C

using

R

-formula reduction.

Soru 3Soru →

Without using a calculator, evaluate the exact value of the trigonometric expression:

\frac{\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ}{\tan 45^\circ + \tan^2 60^\circ}

\frac{1}{4}

\frac{1}{2}

\frac{3}{4}

1

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{4}$

Cevap

\frac{1}{4}

Evaluating each trigonometric ratio using standard special angles (

30^\circ, 45^\circ, 60^\circ

) gives a numerator of

\left(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2} \cdot \frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4} = 1

. The denominator is

\tan 45^\circ + (\tan 60^\circ)^2 = 1 + (\sqrt{3})^2 = 1 + 3 = 4

. Dividing the numerator by the denominator yields

\frac{1}{4}

.

Adım Adım Çözüm

1

Evaluate special angle values for the numerator

\sin 60^\circ = \frac{\sqrt{3}}{2}

,

\cos 30^\circ = \frac{\sqrt{3}}{2}

,

\cos 60^\circ = \frac{1}{2}

, and

\sin 30^\circ = \frac{1}{2}

.

Recall exact surd values for special angles

30^\circ

and

60^\circ

.

2

Simplify the numerator expression

\left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2} \times \frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4} = 1

.

Multiply exact surd values and add fractions.

3

Evaluate special angle values for the denominator

\tan 45^\circ = 1

and

\tan 60^\circ = \sqrt{3}

, so

\tan^2 60^\circ = (\sqrt{3})^2 = 3

.

Square the exact value of

\tan 60^\circ

.

4

Simplify the denominator and divide the numerator by the denominator

\text{Denominator} = 1 + 3 = 4, \text{ so overall value} = \frac{1}{4}$.

Divide numerator result by denominator result.

Anahtar Kavram

Special Angle Trigonometric Ratios and Exact Value Simplification

Soru 4Soru →

A vessel departs from a harbor $H$ and sails $12\text{ km}$ on a bearing of $040^\circ$ to reach point $A$ . From point $A$ , it changes course and sails $5\text{ km}$ on a bearing of $130^\circ$ to reach point $B$ . What is the bearing of point $B$ from harbor $H$ , to the nearest degree?

063^\circ

073^\circ

017^\circ

107^\circ

Cevabı ve açıklamayı göster

Cevap: $063^\circ$

Cevap

063^\circ

(or

063^\circ

to the nearest degree)

The navigation path forms a right-angled triangle at point

A

because the reverse bearing of

H

from

A

(

220^\circ

) differs from the bearing of

B

from

A

(

130^\circ

) by exactly

90^\circ

. Using the right triangle trigonometric ratio

\tan(\angle AHB) = \frac{5}{12}

, the angle

\angle AHB

is found to be approximately

22.62^\circ

. Adding this angle to the initial bearing of

040^\circ

yields

062.62^\circ

, which rounds to

063^\circ

.

Adım Adım Çözüm

1

Determine the back bearing of

H

from

A

and calculate the interior angle at

A

.

Back bearing of

H

from

A = 040^\circ + 180^\circ = 220^\circ

. Interior angle

\angle HAB = 220^\circ - 130^\circ = 90^\circ

.

Knowing the directions of

AH

and

AB

allows us to find the interior angle of

\triangle HAB

at point

A

.

2

Calculate the interior angle

\angle AHB

using right-triangle trigonometry.

\tan(\angle AHB) = \frac{AB}{HA} = \frac{5}{12} \implies \angle AHB = \arctan\left(\frac{5}{12}\right) \approx 22.62^\circ.

Since

\triangle HAB

has a right angle at

A

, tangent relates the opposite side

AB = 5\text{ km}

to the adjacent side

HA = 12\text{ km}

.

3

Calculate the total bearing of

B

from

H

.

\text{Bearing} = 040^\circ + 22.62^\circ = 062.62^\circ \approx 063^\circ.

The line

HB

lies to the right of line

HA

, so the interior angle

\angle AHB

is added to the initial bearing of line

HA

(

040^\circ

).

Anahtar Kavram

Bearings and Right-Angled Triangles

Tahmini Süre:1m 30s

Soru 5Soru →

Two chords $AB$ and $CD$ intersect at a point $P$ inside a circle. If $AP = 4\text{ cm}$ , $PB = 9\text{ cm}$ , and $CP = 3\text{ cm}$ , what is the length of segment $PD$ in centimeters?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The length of segment

PD

is

12\text{ cm}

.

According to the Intersecting Chords Theorem, for two chords intersecting inside a circle at point

P

, the relation

AP \times PB = CP \times PD

holds true. Substituting

AP = 4

,

PB = 9

, and

CP = 3

gives

4 \times 9 = 3 \times PD

, so

36 = 3 \times PD

, which yields

PD = 12\text{ cm}

.

Adım Adım Çözüm

1

Apply the Intersecting Chords Theorem

AP \times PB = CP \times PD

When two chords intersect inside a circle, the product of the segments of one chord equals the product of the segments of the other.

2

Substitute the known values

4 \times 9 = 3 \times PD

, which gives

36 = 3 \times PD

Insert the values

AP = 4\text{ cm}

,

PB = 9\text{ cm}

, and

CP = 3\text{ cm}

.

3

Solve for

PD

PD = 12\text{ cm}

Divide both sides of the equation by 3.

Anahtar Kavram

Intersecting Chords Theorem

Soru 6Soru →

An equilateral triangle has an area of $16\sqrt{3}\text{ cm}^2$ . If a square has the same perimeter as this equilateral triangle, what is the area of the square?

36\text{ cm}^2

48\text{ cm}^2

64\text{ cm}^2

144\text{ cm}^2

Cevabı ve açıklamayı göster

Cevap: $36\text{ cm}^2$

Cevap

The area of the square is

36\text{ cm}^2

.

Using the area formula for an equilateral triangle

\text{Area} = \frac{\sqrt{3}}{4}s^2 = 16\sqrt{3}

, we determine the side length

s = 8\text{ cm}

. The perimeter of the triangle is

3 \times 8\text{ cm} = 24\text{ cm}

. Since the square has an equal perimeter of

24\text{ cm}

, each side of the square measures

24\text{ cm} / 4 = 6\text{ cm}

. The area of the square is therefore

6^2 = 36\text{ cm}^2

.

Adım Adım Çözüm

1

Find the side length of the equilateral triangle.

Side length

s = 8\text{ cm}

.

The area of an equilateral triangle is given by

\text{Area} = \frac{\sqrt{3}}{4}s^2

. Setting

\frac{\sqrt{3}}{4}s^2 = 16\sqrt{3}

yields

s^2 = 64

, so

s = 8\text{ cm}

.

2

Calculate the perimeter of the equilateral triangle.

Perimeter

P = 24\text{ cm}

.

An equilateral triangle has 3 equal sides, so

P = 3 \times 8 = 24\text{ cm}

.

3

Determine the side length of the square.

Square side length

a = 6\text{ cm}

.

The square and triangle have equal perimeters (

24\text{ cm}

). Since a square has 4 equal sides,

a = \frac{24}{4} = 6\text{ cm}

.

4

Calculate the area of the square.

Area

= 36\text{ cm}^2

.

The area of a square is

a^2 = 6^2 = 36\text{ cm}^2

.

Anahtar Kavram

Perimeter and Area of Equilateral Triangles and Squares

Tahmini Süre:1m 30s

Soru 7Soru →

A sector of a circle of radius $21\text{ cm}$ has a total perimeter of $64\text{ cm}$ . Calculate the area of the sector in $\text{cm}^2$ . (Take $\pi = \frac{22}{7}$ )

Cevabı ve açıklamayı göster

Cevap: 231

Cevap

The area of the sector is

231\text{ cm}^2

.

The total perimeter of a sector is given by

P = 2r + l

. Given radius

r = 21\text{ cm}

and perimeter

P = 64\text{ cm}

, the arc length is

l = 64 - 2(21) = 22\text{ cm}

. The area of the sector is calculated using

A = \frac{1}{2} r l = \frac{1}{2} \times 21 \times 22 = 231\text{ cm}^2

.

Adım Adım Çözüm

1

Find the arc length of the sector

The arc length

l = 22\text{ cm}

The total perimeter of a sector includes its arc length plus its two bounding radii (

P = 2r + l

). Subtracting

2r = 42\text{ cm}

from

64\text{ cm}

gives

l = 22\text{ cm}

.

2

Calculate the area of the sector

The area

A = 231\text{ cm}^2

Using the relation between arc length and sector area,

A = \frac{1}{2} r l = \frac{1}{2} \times 21 \times 22 = 231\text{ cm}^2

.

Anahtar Kavram

Perimeter and Area of a Sector of a Circle

Tahmini Süre:1m 30s

Soru 8Soru →

A sports field consists of a central rectangular section of length $100\text{ m}$ bounded on two opposite ends by semicircular regions, each having a radius of $35\text{ m}$ . What is the total area of the sports field? (Take $\pi = \frac{22}{7}$ )

7,350\text{ m}^2

8,925\text{ m}^2

10,850\text{ m}^2

14,700\text{ m}^2

Cevabı ve açıklamayı göster

Cevap: $10,850\text{ m}^2$

Cevap

The total area of the sports field is

10,850\text{ m}^2

.

The sports field is composed of a central rectangle of dimensions

100\text{ m}

by

70\text{ m}

(since width is equal to the diameter

2 \times 35\text{ m} = 70\text{ m}

) and two semicircular ends of radius

35\text{ m}

. The rectangular area is

100 \times 70 = 7,000\text{ m}^2

. The two semicircles join to make one full circle with an area of

\frac{22}{7} \times 35^2 = 3,850\text{ m}^2

. Adding both parts gives

7,000 + 3,850 = 10,850\text{ m}^2

.

Adım Adım Çözüm

1

Calculate the width of the rectangular region.

Width =

2 \times r = 2 \times 35\text{ m} = 70\text{ m}

.

The diameter of the semicircular ends forms the width of the central rectangle.

2

Calculate the area of the rectangular region.

Area of rectangle =

\text{length} \times \text{width} = 100\text{ m} \times 70\text{ m} = 7,000\text{ m}^2

.

Formula for the area of a rectangle is

\text{length} \times \text{width}

.

3

Calculate the combined area of the two semicircular ends.

Combined area =

\pi r^2 = \frac{22}{7} \times 35 \times 35 = 22 \times 5 \times 35 = 3,850\text{ m}^2

.

Two identical semicircles of radius

r

combine to form one full circle of radius

r

.

4

Add the rectangular area and the combined circular area.

Total Area =

7,000\text{ m}^2 + 3,850\text{ m}^2 = 10,850\text{ m}^2

.

The total area is the sum of the composite plane shapes.

Anahtar Kavram

Area of Composite Plane Figures

Soru 9Soru →

A right pyramid has a square base with a perimeter of $24\text{ cm}$ and a vertical height of $10\text{ cm}$ . What is the volume of the pyramid?

120\text{ cm}^3

360\text{ cm}^3

80\text{ cm}^3

240\text{ cm}^3

Cevabı ve açıklamayı göster

Cevap: $120\text{ cm}^3$

Cevap

The volume of the pyramid is

120\text{ cm}^3

.

The volume of a right pyramid is given by

V = \frac{1}{3} \times \text{Base Area} \times h

. First, find the side length of the square base by dividing the perimeter by 4:

s = 24 / 4 = 6\text{ cm}

. The base area is

s^2 = 36\text{ cm}^2

. Substituting into the volume formula gives

V = \frac{1}{3} \times 36 \times 10 = 120\text{ cm}^3

.

Adım Adım Çözüm

1

Determine the side length of the square base from its perimeter.

Side length

s = \frac{24\text{ cm}}{4} = 6\text{ cm}

.

A square has four equal sides, so perimeter divided by 4 yields the side length.

2

Calculate the area of the square base.

Base Area

A = s^2 = 6^2 = 36\text{ cm}^2

.

The area of a square is given by the side length squared.

3

Apply the volume formula for a right pyramid.

Volume

V = \frac{1}{3} \times A \times h = \frac{1}{3} \times 36\text{ cm}^2 \times 10\text{ cm} = 120\text{ cm}^3

.

The volume of any pyramid is one-third of the base area multiplied by the vertical height.

Anahtar Kavram

Volume of a Right Pyramid

Tahmini Süre:1m 15s

Soru 10Soru →

A solid right triangular prism has a base that is a right-angled triangle with legs of length $6\text{ cm}$ and $8\text{ cm}$ . If the height of the prism is $15\text{ cm}$ , what is the total surface area of the prism, in $\text{cm}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 408

Cevap

The total surface area of the right triangular prism is

408\text{ cm}^2

.

To find the total surface area of a right triangular prism, compute the sum of the areas of its 2 triangular bases and its 3 rectangular side faces. The legs of the right-angled triangle are

6\text{ cm}

and

8\text{ cm}

, so the hypotenuse is

\sqrt{6^2 + 8^2} = 10\text{ cm}

. The combined area of the two bases is

2 \times (\frac{1}{2} \times 6 \times 8) = 48\text{ cm}^2

. The perimeter of the base is

6 + 8 + 10 = 24\text{ cm}

, making the lateral surface area

24 \times 15 = 360\text{ cm}^2

. Adding the base areas and lateral area yields

48 + 360 = 408\text{ cm}^2

.

Adım Adım Çözüm

1

Determine the length of the hypotenuse of the triangular base.

Hypotenuse =

10\text{ cm}

.

The base is a right-angled triangle with legs

6\text{ cm}

and

8\text{ cm}

. By Pythagoras:

c = \sqrt{6^2 + 8^2} = 10\text{ cm}

.

2

Calculate the total area of the two parallel triangular bases.

Base area total =

48\text{ cm}^2

.

The area of one right triangle is

\frac{1}{2} \times 6 \times 8 = 24\text{ cm}^2

, so two bases have an area of

2 \times 24 = 48\text{ cm}^2

.

3

Calculate the lateral surface area of the three rectangular faces.

Lateral surface area =

360\text{ cm}^2

.

The lateral surface area is equal to the perimeter of the base times the height:

(6 + 8 + 10) \times 15 = 24 \times 15 = 360\text{ cm}^2

.

4

Sum the base area total and lateral surface area.

Total Surface Area =

408\text{ cm}^2

.

Total surface area =

48 + 360 = 408\text{ cm}^2

.

Anahtar Kavram

Total Surface Area of a Right Triangular Prism

Soru 11Soru →

A decorative wooden cone has a slant height of $13\text{ cm}$ and a vertical height of $12\text{ cm}$ . What is the volume of the cone in terms of $\pi$ ?

100\pi\text{ cm}^3

300\pi\text{ cm}^3

400\pi\text{ cm}^3

65\pi\text{ cm}^3

Cevabı ve açıklamayı göster

Cevap: $100\pi\text{ cm}^3$

Cevap

The volume of the cone is

100\pi\text{ cm}^3

.

The radius is determined via the Pythagorean relation

r = \sqrt{13^2 - 12^2} = 5\text{ cm}

. Substituting

r = 5\text{ cm}

and

h = 12\text{ cm}

into

V = \frac{1}{3}\pi r^2 h

yields

V = \frac{1}{3}\pi(25)(12) = 100\pi\text{ cm}^3

.

Adım Adım Çözüm

1

Find the base radius of the cone using the Pythagorean theorem.

r = \sqrt{l^2 - h^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\text{ cm}

.

The radius, vertical height, and slant height of a right circular cone form a right-angled triangle where the slant height is the hypotenuse.

2

Calculate the volume of the cone using the formula

V = \frac{1}{3}\pi r^2 h

.

V = \frac{1}{3} \times \pi \times 5^2 \times 12 = \frac{1}{3} \times \pi \times 25 \times 12 = 100\pi\text{ cm}^3

.

The formula for the volume of any right circular cone requires multiplying one-third of the base area by the vertical height.

Anahtar Kavram

Volume of a Right Circular Cone

Soru 12Soru →

A metal plate is initially in the form of a rectangle $ABCD$ measuring $14\text{ cm}$ by $10\text{ cm}$ , where $AB = 14\text{ cm}$ . A semicircular piece with diameter $AB$ is cut out from side $AB$ . On the opposite side $CD$ , an isosceles triangular plate with base $CD$ and height $24\text{ cm}$ is attached externally. Taking $\pi = \frac{22}{7}$ , what is the total area of the resulting plate in $\text{cm}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 231

Cevap

The total area of the resulting plate is

231\text{ cm}^2

.

The net area of the plate is found by starting with the area of the rectangle (

140\text{ cm}^2

), subtracting the area of the semicircular cutout (

77\text{ cm}^2

), and adding the area of the attached isosceles triangle (

168\text{ cm}^2

), resulting in

140 - 77 + 168 = 231\text{ cm}^2

.

Adım Adım Çözüm

1

Calculate the area of the original rectangle ABCD

140 cm²

The area of a rectangle is calculated as

\text{length} \times \text{width} = 14\text{ cm} \times 10\text{ cm} = 140\text{ cm}^2

.

2

Calculate the area of the removed semicircular section

77 cm²

The radius of the semicircle is

r = \frac{14}{2} = 7\text{ cm}

. The area of a semicircle is

\frac{1}{2}\pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = 77\text{ cm}^2

.

3

Calculate the area of the attached triangular section

168 cm²

The area of a triangle is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14\text{ cm} \times 24\text{ cm} = 168\text{ cm}^2

.

4

Combine the area components to determine the final net area

231 cm²

Subtract the removed semicircular area from the rectangle's area and add the attached triangle's area:

140 - 77 + 168 = 231\text{ cm}^2

.

Anahtar Kavram

Perimeter and Area of Composite Plane Figures

Soru 13Soru →

A boat sails $10\text{ km}$ from a port $P$ on a bearing of $040^\circ$ to a point $Q$ . From $Q$ , it changes direction and sails $10\sqrt{3}\text{ km}$ on a bearing of $130^\circ$ to reach a point $R$ . What is the bearing of port $P$ from point $R$ ?

280^\circ

100^\circ

250^\circ

160^\circ

Cevabı ve açıklamayı göster

Cevap: $280^\circ$

Cevap

The bearing of port P from point R is

280^\circ

.

By drawing North reference lines at P, Q, and R, the back bearing of P from Q is

040^\circ + 180^\circ = 220^\circ

. The bearing of R from Q is

130^\circ

, creating an interior right angle of

220^\circ - 130^\circ = 90^\circ

at Q. Using right-triangle trigonometry,

\tan(\angle QPR) = \frac{10\sqrt{3}}{10} = \sqrt{3}

, giving

\angle QPR = 60^\circ

. Adding this to the initial bearing of

040^\circ

gives the bearing of R from P as

100^\circ

. Finally, adding

180^\circ

gives the reverse bearing of P from R as

280^\circ

.

Adım Adım Çözüm

1

Determine the interior angle

\angle PQR

at vertex

Q

\angle PQR = 90^\circ

The back bearing of P from Q is

040^\circ + 180^\circ = 220^\circ

. The bearing of R from Q is

130^\circ

. The interior angle between QP and QR is

220^\circ - 130^\circ = 90^\circ

.

2

Calculate the angle

\angle QPR

inside right-angled triangle

PQR

\angle QPR = 60^\circ

Since triangle

PQR

is right-angled at

Q

,

\tan(\angle QPR) = \frac{\text{opposite}}{\text{adjacent}} = \frac{QR}{PQ} = \frac{10\sqrt{3}}{10} = \sqrt{3}

. Therefore,

\angle QPR = \arctan(\sqrt{3}) = 60^\circ

.

3

Find the forward bearing of point

R

from port

P

Bearing of

R

from

P = 100^\circ

The bearing of Q from P is

040^\circ

. Since R lies to the right (clockwise) of segment PQ, add

\angle QPR = 60^\circ

to

040^\circ

:

040^\circ + 60^\circ = 100^\circ

.

4

Compute the back bearing of port

P

from point

R

Bearing of

P

from

R = 280^\circ

The back bearing is obtained by adding

180^\circ

to the forward bearing from P to R:

100^\circ + 180^\circ = 280^\circ

.

Anahtar Kavram

Three-point bearing calculations using right-triangle trigonometry and back bearings

Tahmini Süre:2m 30s

Soru 14Soru →

A line $L_1$ is given by the equation $3x + 4y - 24 = 0$ , intersecting the x-axis at point $A$ and the y-axis at point $B$ . A second line $L_2$ has the equation $4x - 3y + k = 0$ , where $k > 0$ . If the perpendicular distance from the midpoint of the line segment $AB$ to $L_2$ is $5$ units, what is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

The value of

k

is

18

.

To find

k

, first determine the intercepts of

L_1

: setting

y=0

gives

A(8, 0)

and setting

x=0

gives

B(0, 6)

. The midpoint

M

of segment

AB

is

\left(\frac{8+0}{2}, \frac{0+6}{2}\right) = (4, 3)

. Using the perpendicular distance formula

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

for point

M(4,3)

and line

4x - 3y + k = 0

, we get

d = \frac{|4(4) - 3(3) + k|}{\sqrt{4^2 + (-3)^2}} = \frac{|7 + k|}{5}

. Setting

d = 5

gives

|7 + k| = 25

. Since

k > 0

, solving

7 + k = 25

gives the correct value

k = 18

.

Adım Adım Çözüm

1

Determine the coordinates of points A and B

A = (8, 0)

and

B = (0, 6)

Setting

y = 0

in

3x + 4y - 24 = 0

yields

3x = 24 \implies x = 8

. Setting

x = 0

yields

4y = 24 \implies y = 6

.

2

Calculate the midpoint M of segment AB

M = (4, 3)

Using the midpoint formula

M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{8 + 0}{2}, \frac{0 + 6}{2}\right) = (4, 3)

.

3

Set up the perpendicular distance equation from M(4,3) to line L₂

d = \frac{|7 + k|}{5}

Applying the distance formula

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

gives

d = \frac{|4(4) - 3(3) + k|}{\sqrt{4^2 + (-3)^2}} = \frac{|16 - 9 + k|}{\sqrt{25}} = \frac{|7 + k|}{5}

.

4

Solve for k given d = 5 and k > 0

k = 18

Equating distance to

5

gives

\frac{|7 + k|}{5} = 5 \implies |7 + k| = 25

. Since

k > 0

,

7 + k = 25

, which yields

k = 18

.

Anahtar Kavram

Perpendicular Distance from a Point to a Straight Line

Tahmini Süre:2m 30s

Soru 15Soru →

Simplify the trigonometric expression $\frac{\tan 60^\circ + \sin 45^\circ}{\cos 45^\circ}$ to its exact surd form.

\sqrt{6} + 1

\frac{\sqrt{10}}{2}

\frac{3 + \sqrt{6}}{3}

\sqrt{3} + 1

Cevabı ve açıklamayı göster

Cevap: $\sqrt{6} + 1$

Cevap

\sqrt{6} + 1

Substituting the exact values gives

\tan 60^\circ = \sqrt{3}

,

\sin 45^\circ = \frac{1}{\sqrt{2}}

, and

\cos 45^\circ = \frac{1}{\sqrt{2}}

. Simplifying

\frac{\sqrt{3} + 1/\sqrt{2}}{1/\sqrt{2}}

gives

\sqrt{3} \cdot \sqrt{2} + 1 = \sqrt{6} + 1

.

Adım Adım Çözüm

1

Substitute the exact trigonometric values for the special angles

\tan 60^\circ = \sqrt{3}

,

\sin 45^\circ = \frac{1}{\sqrt{2}}

, and

\cos 45^\circ = \frac{1}{\sqrt{2}}

Special angle values must be expressed in exact surd form.

2

Set up the fractional expression

\frac{\sqrt{3} + \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}

Replace each ratio with its exact surd equivalent.

3

Divide numerator terms by the denominator

\frac{\sqrt{3}}{\frac{1}{\sqrt{2}}} + \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{3} \cdot \sqrt{2} + 1 = \sqrt{6} + 1

Dividing by a fraction is equivalent to multiplying by its reciprocal.

Anahtar Kavram

Evaluation of Special Angle Trigonometric Ratios and Simplification of Surds

Soru 16Soru →

Find the sum, in degrees, of all solutions to the trigonometric equation $\sqrt{3}\tan(2x) = 3$ in the interval $0^\circ \le x \le 180^\circ$ .

Cevabı ve açıklamayı göster

Cevap: 150

Cevap

The sum of all solutions to the equation in the given interval is 150 degrees.

Isolating

\tan(2x)

gives

\sqrt{3}

. For

0^\circ \le 2x \le 360^\circ

,

\tan(2x) = \sqrt{3}

yields solutions at

2x = 60^\circ

and

2x = 240^\circ

. Dividing by 2 gives

x = 30^\circ

and

x = 120^\circ

. Adding these solutions yields

30^\circ + 120^\circ = 150^\circ

.

Adım Adım Çözüm

1

Isolate the trigonometric function

\tan(2x) = \frac{3}{\sqrt{3}} = \sqrt{3}

Dividing both sides by \sqrt{3} simplifies the expression to a standard special angle ratio.

2

Determine the domain for the argument

2x

Since

0^\circ \le x \le 180^\circ

, multiplying the inequality by 2 gives

0^\circ \le 2x \le 360^\circ

.

This establishes the range of angles to search for

2x

within one complete turn.

3

Find all values of

2x

where tangent equals

\sqrt{3}

2x = 60^\circ

(1st quadrant) and

2x = 180^\circ + 60^\circ = 240^\circ

(3rd quadrant)

The tangent function is positive in Quadrants I and III with a reference angle of

60^\circ

.

4

Solve for

x

x = \frac{60^\circ}{2} = 30^\circ

and

x = \frac{240^\circ}{2} = 120^\circ

Dividing each angle by 2 yields the values of

x

lying within the domain

0^\circ \le x \le 180^\circ

.

5

Calculate the sum of the solutions

30^\circ + 120^\circ = 150^\circ

The question specifically requests the sum of all valid solutions.

Anahtar Kavram

Solving trigonometric equations using reference angles and domain transformation

Tahmini Süre:2m 0s

Soru 17Soru →

In $\Delta ABC$ , side $a = 6\text{ cm}$ , side $b = 10\text{ cm}$ , and the included angle $\angle C = 120^\circ$ . What is the length of side $c$ ?

14\text{ cm}

2\sqrt{19}\text{ cm}

2\sqrt{34}\text{ cm}

16\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $14\text{ cm}$

Cevap

The length of side

c

is

14\text{ cm}

.

According to the Cosine Rule

c^2 = a^2 + b^2 - 2ab \cos C

, substituting the given values

a = 6

,

b = 10

, and

\cos 120^\circ = -\frac{1}{2}

yields

c^2 = 36 + 100 - 2(6)(10)\left(-\frac{1}{2}\right) = 136 + 60 = 196

. Taking the positive square root gives

c = 14\text{ cm}

.

Adım Adım Çözüm

1

Identify the given values and state the relevant Cosine Rule formula

Given:

a = 6\text{ cm}

,

b = 10\text{ cm}

,

\angle C = 120^\circ

. Formula:

c^2 = a^2 + b^2 - 2ab \cos C

.

Since two sides and the included angle (SAS configuration) are known, the Cosine Rule must be used to find the third side.

2

Evaluate

\cos 120^\circ

and substitute all values into the formula

\cos 120^\circ = -\frac{1}{2}

. Thus,

c^2 = 6^2 + 10^2 - 2(6)(10)\left(-\frac{1}{2}\right)

.

Cosine of an obtuse angle in the second quadrant is negative.

3

Simplify the algebraic expression

c^2 = 36 + 100 + 60 = 196

.

Multiplying

-2(60)\left(-\frac{1}{2}\right)

yields

+60

.

4

Take the principal square root to solve for

c

c = \sqrt{196} = 14\text{ cm}

.

Length must be a positive real number.

Anahtar Kavram

Cosine Rule for finding an unknown side in SAS triangle configurations

Soru 18Soru →

Evaluate the trigonometric expression $\tan 60^\circ \times \sin 60^\circ$ . What is the exact simplified value of the expression?

\frac{3}{2}

\frac{\sqrt{3}}{2}

3

\frac{\sqrt{3}}{4}

Cevabı ve açıklamayı göster

Cevap: $\frac{3}{2}$

Cevap

The exact simplified value of the expression is

\frac{3}{2}

.

Substituting the exact surd values of the special angles yields

\tan 60^\circ = \sqrt{3}

and

\sin 60^\circ = \frac{\sqrt{3}}{2}

. Multiplying these gives

\sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{3}{2}

, which is the correct exact value.

Adım Adım Çözüm

1

Identify the values of the special trigonometric angles.

\tan 60^\circ = \sqrt{3}

and

\sin 60^\circ = \frac{\sqrt{3}}{2}

.

Special angle values must be known in exact surd form.

2

Multiply the two trigonometric ratios together.

\tan 60^\circ \times \sin 60^\circ = \sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{(\sqrt{3} \times \sqrt{3})}{2}

.

Substitute the special angle values into the product expression.

3

Simplify the radical expression in the numerator.

\frac{3}{2}

.

The product of

\sqrt{3} \times \sqrt{3}

equals

3

.

Anahtar Kavram

Evaluation of Special Angle Trigonometric Values

Tahmini Süre:45s

Soru 19Soru →

If $x$ is an acute angle such that $\sin x - \cos x = \frac{1}{\sqrt{5}}$ , what is the exact value of $\tan x + \cot x$ ?

\frac{5}{2}

\frac{5}{4}

\frac{2}{5}

\frac{4}{5}

Cevabı ve açıklamayı göster

Cevap: $\frac{5}{2}$

Cevap

The exact value of

\tan x + \cot x

is

\frac{5}{2}

.

Squaring both sides of

\sin x - \cos x = \frac{1}{\sqrt{5}}

yields

1 - 2\sin x \cos x = \frac{1}{5}

, which simplifies to

\sin x \cos x = \frac{2}{5}

. Expressing

\tan x + \cot x

in terms of sine and cosine gives

\frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}

. Substituting

\frac{2}{5}

gives

\frac{1}{2/5} = \frac{5}{2}

.

Adım Adım Çözüm

1

Square both sides of the given equation

\sin x - \cos x = \frac{1}{\sqrt{5}}

(\sin x - \cos x)^2 = \left(\frac{1}{\sqrt{5}}\right)^2 \implies \sin^2 x - 2\sin x \cos x + \cos^2 x = \frac{1}{5}

Squaring enables the use of the Pythagorean trigonometric identity.

2

Apply the fundamental identity

\sin^2 x + \cos^2 x = 1

to isolate

\sin x \cos x

1 - 2\sin x \cos x = \frac{1}{5} \implies 2\sin x \cos x = 1 - \frac{1}{5} = \frac{4}{5} \implies \sin x \cos x = \frac{2}{5}

This determines the value of the product of sine and cosine.

3

Rewrite

\tan x + \cot x

using quotient identities

\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}

Expressing tangent and cotangent with a common denominator simplifies the expression into a reciprocal.

4

Substitute

\sin x \cos x = \frac{2}{5}

into the simplified expression

\tan x + \cot x = \frac{1}{2/5} = \frac{5}{2}

Inverting the fraction gives the final numerical answer.

Anahtar Kavram

Pythagorean and Quotient Trigonometric Identities

Tahmini Süre:2m 0s

Soru 20Soru →

Given that $\sin \theta = \frac{5}{13}$ , where $\theta$ is an acute angle, evaluate the value of $13 \cos \theta - 12 \tan \theta$ . What is the numerical value?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The numerical value of the expression

13 \cos \theta - 12 \tan \theta

is 7.

For an acute angle

\theta

with

\sin \theta = \frac{5}{13}

, the corresponding right triangle has an opposite side of 5, a hypotenuse of 13, and an adjacent side of

\sqrt{13^2 - 5^2} = 12

. Therefore,

\cos \theta = \frac{12}{13}

and

\tan \theta = \frac{5}{12}

. Evaluating

13 \cos \theta - 12 \tan \theta

yields

13\left(\frac{12}{13}\right) - 12\left(\frac{5}{12}\right) = 12 - 5 = 7

.

Adım Adım Çözüm

1

Determine

\cos \theta

using the right triangle ratio or Pythagorean identity.

\cos \theta = \frac{12}{13}

Since

\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}

, the adjacent side is

\sqrt{13^2 - 5^2} = 12

. Because

\theta

is acute,

\cos \theta

is positive.

2

Determine

\tan \theta

using the ratio of opposite to adjacent sides.

\tan \tan \theta = \frac{5}{12}

\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{12}$.

3

Substitute the evaluated ratios into

13 \cos \theta - 12 \tan \theta

and simplify.

13\left(\frac{12}{13}\right) - 12\left(\frac{5}{12}\right) = 12 - 5 = 7

Multiplying clears the denominators, leaving

12 - 5 = 7

.

Anahtar Kavram

Basic Trigonometric Ratios and Pythagorean Triples

Tahmini Süre:1m 15s

Geometry and Trigonometry

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