Geometry and Trigonometry Practice Questions — JAMB UTME — Page 9

Question 161Question →

The sum of all the interior angles of a convex polygon, except for one, is $2190^\circ$ . If the measure of the remaining interior angle is strictly less than $180^\circ$ , how many sides does the polygon have?

13

14

15

16

Show answer & explanation

Answer: 15

Answer

The polygon has 15 sides.

The total sum of interior angles of an

n

-sided polygon is

(n - 2) \times 180^\circ

. Given that all angles except one sum to

2190^\circ

and the missing angle is between

0^\circ

and

180^\circ

, the total sum must be the smallest multiple of

180^\circ

greater than

2190^\circ

. The next multiple of

180^\circ

after

2190^\circ

is

2340^\circ

. Dividing

2340^\circ

by

180^\circ

gives

13

, so

n - 2 = 13

, which yields

n = 15

. The missing angle is

2340^\circ - 2190^\circ = 150^\circ

, which is valid.

Step-by-Step Solution

1

State the formula for the sum of interior angles of an

n

-sided convex polygon.

S = (n - 2) \times 180^\circ

, where

n

is an integer representing the number of sides (

n \ge 3

).

The sum of interior angles of any convex polygon is determined by dividing it into

(n - 2)

triangles.

2

Set up an inequality using the given partial sum and the constraint on the remaining interior angle.

Let the missing interior angle be

x

. Then

(n - 2) \times 180^\circ = 2190^\circ + x

, with

0^\circ < x < 180^\circ

.

Since the polygon is convex, every interior angle is non-reflex and greater than

0^\circ

.

3

Solve for the bounds of

(n - 2)

.

Dividing

2190^\circ < (n - 2) \times 180^\circ < 2190^\circ + 180^\circ

by

180^\circ

gives

12.166... < n - 2 < 13.166...

. Since

n - 2

must be an integer,

n - 2 = 13

.

The total interior angle sum must be an exact integer multiple of

180^\circ

.

4

Calculate

n

and verify the value of

x

.

n = 13 + 2 = 15

. The total sum is

13 \times 180^\circ = 2340^\circ

, making the remaining angle

x = 2340^\circ - 2190^\circ = 150^\circ

.

The remaining angle

150^\circ

satisfies

0^\circ < 150^\circ < 180^\circ

, confirming

n = 15

is correct.

Key Concept

Sum of Interior Angles of Convex Polygons

Question 162Question →

What is the exact numerical value of the trigonometric expression

\frac{12 \sin 30^\circ \cos 30^\circ \tan 60^\circ}{\sin^2 45^\circ + \cos^2 45^\circ + \tan^2 60^\circ}

?

Show answer & explanation

Answer: 2.25

Answer

The exact numerical value of the expression is 2.25.

Substituting the exact values

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

,

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

, and

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

into the numerator gives

12 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} \times \sqrt{3} = 9

. Using the Pythagorean identity

\sin^2 45^\circ + \cos^2 45^\circ = 1

and

\tan^2 60^\circ = 3

, the denominator evaluates to

1 + 3 = 4

. Dividing 9 by 4 yields the exact value of 2.25.

Step-by-Step Solution

1

Evaluate the trigonometric ratios for special angles and apply trigonometric identities.

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

,

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

,

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

, and

\sin^2 45^\circ + \cos^2 45^\circ = 1

.

Standard special angle values and the fundamental Pythagorean identity

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

simplify the expression.

2

Substitute these values into the numerator of the expression.

Numerator

= 12 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} \times \sqrt{3} = 9

.

Multiplying the terms:

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

, and

12 \times \frac{1}{4} \times 3 = 9

.

3

Substitute these values into the denominator of the expression.

Denominator

= 1 + (\sqrt{3})^2 = 1 + 3 = 4

.

The sum

\sin^2 45^\circ + \cos^2 45^\circ = 1

combined with

\tan^2 60^\circ = 3

equals 4.

4

Divide the calculated numerator by the denominator.

\frac{9}{4} = 2.25

.

Dividing 9 by 4 gives the final decimal value 2.25.

Key Concept

Evaluation of trigonometric expressions using special angles (30°, 45°, 60°) and fundamental identities

Question 163Question →

A straight line $L_1$ has the equation $3x - 4y + 5 = 0$ . A second line $L_2$ is parallel to $L_1$ and passes through the point $(6, 1)$ . What is the perpendicular distance between lines $L_1$ and $L_2$ ?

Show answer & explanation

Answer: 3.8

Answer

The perpendicular distance between lines

L_1

and

L_2

is

3.8

units.

The perpendicular distance between parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

. Line

L_2

is parallel to

3x - 4y + 5 = 0

, so its equation is

3x - 4y + C = 0

. Substituting

(6, 1)

gives

3(6) - 4(1) + C = 0

, leading to

C = -14

. Substituting

C_1 = 5

and

C_2 = -14

into the distance formula gives

d = \frac{|5 - (-14)|}{\sqrt{3^2 + (-4)^2}} = \frac{19}{5} = 3.8

.

Step-by-Step Solution

1

Determine the equation of line

L_2

The equation of

L_2

is

3x - 4y - 14 = 0

Lines parallel to

3x - 4y + 5 = 0

have the form

3x - 4y + C = 0

. Substituting the point

(6, 1)

gives

3(6) - 4(1) + C = 0 \implies C = -14

.

2

Apply the parallel line distance formula

d = \frac{|5 - (-14)|}{\sqrt{3^2 + (-4)^2}} = \frac{19}{5}

The perpendicular distance between parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

.

3

Convert fraction to decimal form

3.8

Dividing

19

by

5

yields

3.8

.

Key Concept

Perpendicular Distance Between Parallel Lines

Question 164Question →

A convex polygon with $n$ sides has a total interior angle sum of $1440^\circ$ . If $(n - 4)$ of its interior angles each measure $150^\circ$ , and the remaining four interior angles are equal in measure, what is the measure of one of the remaining interior angles in degrees?

Show answer & explanation

Answer: 135

Answer

The measure of one of the remaining interior angles is

135^\circ

.

Using the interior angle sum formula

(n - 2) \times 180^\circ = 1440^\circ

, we find

n - 2 = 8

, which means the polygon has

n = 10

sides. The number of angles measuring

150^\circ

is

10 - 4 = 6

. Their total measure is

6 \times 150^\circ = 900^\circ

. The sum of the remaining four equal angles is

1440^\circ - 900^\circ = 540^\circ

. Dividing

540^\circ

by 4 gives

135^\circ

for each remaining interior angle.

Step-by-Step Solution

1

Calculate the total number of sides

n

of the convex polygon.

n = 10

The sum of interior angles of an

n

-sided polygon is given by

(n - 2) \times 180^\circ

. Setting

(n - 2) \times 180^\circ = 1440^\circ

gives

n - 2 = 8

, so

n = 10

.

2

Determine the number of interior angles that measure

150^\circ

each and find their combined sum.

6 angles totaling

900^\circ

There are

(n - 4) = 10 - 4 = 6

interior angles of

150^\circ

each. Their sum is

6 \times 150^\circ = 900^\circ

.

3

Calculate the sum of the remaining four equal interior angles.

540^\circ

Subtracting the sum of the known angles from the total interior angle sum yields

1440^\circ - 900^\circ = 540^\circ

.

4

Find the measure of one of the remaining four equal angles.

135^\circ

Dividing the remaining sum equally among the 4 angles gives

540^\circ / 4 = 135^\circ

.

Key Concept

Polygon interior angle sum theorem

Estimated Time:1m 30s

Question 165Question →

If $\theta$ is an acute angle such that $\sin \theta = \frac{5}{13}$ , what is the exact numerical value of $169(\sin^2 \theta - \cos^2 \theta)$ ?

Show answer & explanation

Answer: -119

Answer

The exact numerical value of the expression is

-119

.

Using the Pythagorean identity

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

, we find

\cos^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}

. Then

\sin^2 \theta - \cos^2 \theta = \frac{25 - 144}{169} = -\frac{119}{169}

. Multiplying by

169

gives

-119

.

Step-by-Step Solution

1

Calculate

\cos^2 \theta

using the Pythagorean trigonometric identity.

\cos^2 \theta = \frac{144}{169}

Since

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

, subtracting

\sin^2 \theta = \frac{25}{169}

from

1

yields

\frac{144}{169}

.

2

Compute the difference

\sin^2 \theta - \cos^2 \theta

.

\sin^2 \theta - \cos^2 \theta = -\frac{119}{169}

Subtracting

\frac{144}{169}

from

\frac{25}{169}

gives

-\frac{119}{169}

.

3

Scale the difference by

169

.

169 \times \left(-\frac{119}{169}\right) = -119

Multiplying

-\frac{119}{169}

by

169

cancels the denominator, leaving

-119

.

Key Concept

Pythagorean Trigonometric Identity

Question 166Question →

A trapezium has an area of $180\text{ cm}^2$ and a perpendicular height of $12\text{ cm}$ . If the lengths of its two parallel sides are in the ratio $2:3$ , calculate the length, in $\text{cm}$ , of the longer parallel side.

Show answer & explanation

Answer: 18

Answer

The length of the longer parallel side is

18\text{ cm}

.

By representing the parallel sides as

2x

and

3x

, the trapezium area formula

A = \frac{1}{2}(a + b)h

gives

180 = \frac{1}{2}(2x + 3x)(12) = 30x

. Solving for

x

yields

x = 6

. Therefore, the longer parallel side is

3(6) = 18\text{ cm}

.

Step-by-Step Solution

1

Express the parallel sides algebraically from the ratio

2:3

.

Let the shorter parallel side

a = 2x

and the longer parallel side

b = 3x

.

Using a common multiplier

x

preserves the given side ratio.

2

Substitute the expressions and known values into the trapezium area formula.

180 = \frac{1}{2}(2x + 3x) \times 12

The area

A

of a trapezium is given by

A = \frac{1}{2}(a + b)h

.

3

Solve for the variable

x

.

180 = 6 \times 5x \implies 30x = 180 \implies x = 6

Simplifying

\frac{1}{2} \times 12 = 6

and multiplying by

(2x + 3x) = 5x

.

4

Calculate the length of the longer parallel side.

Longer side

= 3x = 3 \times 6 = 18\text{ cm}

The longer side corresponds to the

3x

term in the ratio.

Key Concept

Area of a trapezium involving algebraic ratio problem solving

Question 167Question →

A point $P(x, y)$ moves in the Cartesian plane such that it maintains a constant distance of $10\text{ units}$ from a fixed point $C(2, -3)$ . If the locus of $P$ intersects the vertical line $x = 8$ at two points $A$ and $B$ , what is the distance between $A$ and $B$ ?

Show answer & explanation

Answer: 16

Answer

The distance between the intersection points A and B is 16 units.

The locus of point P moving at a constant distance of 10 units from C(2, -3) forms a circle

(x - 2)^2 + (y + 3)^2 = 100

. Substituting

x = 8

yields

(y + 3)^2 = 64

, giving

y = 5

and

y = -11

. The distance between the two points (8, 5) and (8, -11) along the vertical line

x = 8

is

5 - (-11) = 16

units.

Step-by-Step Solution

1

Determine the equation representing the locus of point P

(x - 2)^2 + (y + 3)^2 = 100

The locus of a point moving at a fixed distance from a fixed point is a circle with equation

(x - h)^2 + (y - k)^2 = r^2

.

2

Substitute

x = 8

into the locus equation to find the

y

-coordinates of the intersection points

(8 - 2)^2 + (y + 3)^2 = 100 \implies 36 + (y + 3)^2 = 100 \implies (y + 3)^2 = 64

The intersection points lie on both the locus circle and the vertical line

x = 8

.

3

Solve for the two possible values of

y

y + 3 = \pm 8 \implies y_1 = 5

and

y_2 = -11

Taking the square root gives both positive and negative solutions for the vertical coordinate.

4

Calculate the vertical distance between points

A(8, 5)

and

B(8, -11)

\text{Distance} = 5 - (-11) = 16\text{ units}

Since both points have the same

x

-coordinate (

x = 8

), the distance is simply the absolute difference between their

y

-coordinates.

Key Concept

Locus of a point at a constant distance from a fixed point (Circle)

Question 168Question →

How many distinct solutions exist for the trigonometric equation $2\cos 2\theta = 1$ within the interval $0^\circ \le \theta \le 360^\circ$ ?

1

2

4

8

Show answer & explanation

Answer: 4

Answer

4 distinct solutions

The equation

\cos 2\theta = \frac{1}{2}

requires finding all angles whose cosine is

\frac{1}{2}

. Because the angle argument is

2\theta

, as

\theta

completes one full rotation (

0^\circ

to

360^\circ

),

2\theta

completes two full rotations (

0^\circ

to

720^\circ

). In two rotations, the cosine function takes the value

+\frac{1}{2}

exactly four times (twice per rotation), leading to four distinct solutions for

\theta

.

Step-by-Step Solution

1

Isolate the trigonometric function in the equation

\cos 2\theta = \frac{1}{2}

Dividing both sides by 2 isolates the cosine term.

2

Determine the working interval for the multiple angle

2\theta

0^\circ \le 2\theta \le 720^\circ

Since

0^\circ \le \theta \le 360^\circ

, multiplying the inequality by 2 gives the domain for

2\theta

.

3

Find all values of

2\theta

where cosine is positive

\frac{1}{2}

in the domain

[0^\circ, 720^\circ]

2\theta = 60^\circ, 300^\circ, 420^\circ, 660^\circ

Cosine is positive in Quadrants I and IV. In the first revolution (

0^\circ \text{ to } 360^\circ

),

2\theta = 60^\circ

and

360^\circ - 60^\circ = 300^\circ

. Adding

360^\circ

for the second revolution gives

420^\circ

and

660^\circ

.

4

Solve for

\theta

by dividing each angle by 2

\theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ

Dividing all four values of

2\theta

by 2 gives four distinct values of

\theta

within

[0^\circ, 360^\circ]

.

Key Concept

Solving trigonometric equations with multiple angles over a specified domain

Question 169Question →

In $\Delta ABC$ , the side lengths are given as $a = 7\text{ cm}$ , $b = 5\text{ cm}$ , and $c = 3\text{ cm}$ . What is the measure of angle $A$ in degrees?

Show answer & explanation

Answer: 120

Answer

The measure of angle

A

is

120^\circ

.

Using the Cosine Rule formula

\cos A = \frac{b^2 + c^2 - a^2}{2bc}

, substituting

a = 7

,

b = 5

, and

c = 3

yields

\cos A = \frac{25 + 9 - 49}{30} = -\frac{1}{2}

. The inverse cosine of

-\frac{1}{2}

gives an obtuse angle of

120^\circ

.

Step-by-Step Solution

1

Apply the Cosine Rule for an unknown angle in terms of the three sides

\cos A = \frac{b^2 + c^2 - a^2}{2bc}

When all three side lengths of a non-right triangle are given (SSS), the Cosine Rule is required to solve for any internal angle.

2

Substitute

a = 7

,

b = 5

, and

c = 3

into the Cosine Rule formula and evaluate

\cos A = \frac{25 + 9 - 49}{2 \times 5 \times 3} = \frac{-15}{30} = -0.5

Evaluating the terms in the numerator and denominator simplifies the expression for

\cos A

.

3

Calculate the inverse cosine of

-0.5

to find angle

A

A = 120^\circ

Since the cosine value is negative, angle

A

is obtuse and lies in the second quadrant (

90^\circ < A < 180^\circ

).

Key Concept

Using the Cosine Rule with three side lengths (SSS) to find an obtuse interior angle

Estimated Time:1m 30s

Question 170Question →

Points $P(k, 2)$ and $Q(3, 8)$ lie on a straight line $L_1$ . If $L_1$ is perpendicular to the line $L_2$ given by $4x + 3y - 12 = 0$ , what is the value of $k$ ?

Show answer & explanation

Answer: -5

Answer

The value of

k

is

-5

.

To determine

k

, the gradient of

L_2

(

4x + 3y - 12 = 0

) is found to be

-\frac{4}{3}

. Using the perpendicularity rule

m_1 \cdot m_2 = -1

, the gradient of

L_1

is

\frac{3}{4}

. Equating this to the slope formula

\frac{8 - 2}{3 - k}

gives

\frac{6}{3 - k} = \frac{3}{4}

, which simplifies to

k = -5

.

Step-by-Step Solution

1

Find the gradient

m_2

of line

L_2

m_2 = -\frac{4}{3}

Converting

4x + 3y - 12 = 0

to

y = mx + c

form gives

y = -\frac{4}{3}x + 4

.

2

Apply the perpendicular line condition to find

m_1

m_1 = \frac{3}{4}

Perpendicular lines have negative reciprocal gradients (

m_1 \cdot m_2 = -1

).

3

Express the gradient

m_1

using the coordinates of

P

and

Q

m_1 = \frac{6}{3 - k}

Using the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

for points

P(k, 2)

and

Q(3, 8)

.

4

Solve for

k

k = -5

Equating

\frac{6}{3 - k} = \frac{3}{4}

yields

3(3 - k) = 24 \implies 9 - 3k = 24 \implies k = -5

.

Key Concept

Perpendicular Lines and Gradient Formula

Question 171Question →

What is the equation of the locus of a point $P(x, y)$ that is always equidistant from the point $(0, 4)$ and the line $y = -4$ ?

Show answer & explanation

Answer: x^2 = 16y; x^2 - 16y = 0; x^2-16y=0; y = x^2/16; y = \frac{x^2}{16}; x^{2}=16y; x^{2}-16y=0

Answer

The equation of the locus is

x^2 = 16y

(or

x^2 - 16y = 0

).

Equating the distance from

P(x, y)

to

(0, 4)

, which is

\sqrt{x^2 + (y-4)^2}

, to the perpendicular distance to the line

y = -4

, which is

|y + 4|

, and squaring both sides gives

x^2 + (y-4)^2 = (y+4)^2

. Expanding yields

x^2 + y^2 - 8y + 16 = y^2 + 8y + 16

, which simplifies directly to

x^2 = 16y

or

x^2 - 16y = 0

.

Step-by-Step Solution

1

Formulate the distance expressions from point

P(x, y)

to the given point

(0, 4)

and line

y = -4

.

Distance to

(0, 4) = \sqrt{(x - 0)^2 + (y - 4)^2}

. Perpendicular distance to

y = -4

is

|y - (-4)| = |y + 4|

.

By definition of geometric locus, the distance from

P(x, y)

to the fixed point must equal its distance to the fixed line.

2

Equate the two distance expressions.

\sqrt{x^2 + (y - 4)^2} = |y + 4|

The point

P(x, y)

is equidistant from both geometric entities.

3

Square both sides of the equation to clear the square root and absolute value.

x^2 + (y - 4)^2 = (y + 4)^2

Squaring eliminates radical and absolute value signs while preserving algebraic equality.

4

Expand both squared binomial expressions and simplify.

x^2 + y^2 - 8y + 16 = y^2 + 8y + 16 \implies x^2 = 16y

Subtracting

y^2 + 16

from both sides leaves

x^2 - 8y = 8y

, which simplifies to

x^2 = 16y

.

Key Concept

The locus of points equidistant from a fixed point (focus) and a fixed straight line (directrix) forms a parabola.

Estimated Time:1m 30s

Question 172Question →

A right-angled triangle has a perimeter of $60\text{ cm}$ and a hypotenuse of length $25\text{ cm}$ . What is the area of the triangle in $\text{cm}^2$ ?

Show answer & explanation

Answer: 150

Answer

150

Let the perpendicular sides of the right-angled triangle be

a

and

b

, and the hypotenuse be

c = 25\text{ cm}

. From the perimeter,

a + b + 25 = 60

, so

a + b = 35\text{ cm}

. By the Pythagorean theorem,

a^2 + b^2 = 25^2 = 625

. Squaring both sides of

a + b = 35

yields

(a + b)^2 = a^2 + b^2 + 2ab = 35^2 = 1225

. Substituting

a^2 + b^2 = 625

gives

625 + 2ab = 1225 \implies 2ab = 600 \implies ab = 300

. The area of the right-angled triangle is

\frac{1}{2}ab = \frac{1}{2} \times 300 = 150\text{ cm}^2

.

Step-by-Step Solution

1

Determine the sum of the two legs from the perimeter

a + b = 35\text{ cm}

The perimeter of the triangle is

a + b + c = 60\text{ cm}

, where the hypotenuse

c = 25\text{ cm}

.

2

Use the Pythagorean theorem for the sum of squares of the legs

a^2 + b^2 = 625

In any right-angled triangle with hypotenuse

25\text{ cm}

,

a^2 + b^2 = 25^2 = 625

.

3

Expand

(a + b)^2

to find the product of the legs

ab

1225 = 625 + 2ab \implies 2ab = 600 \implies ab = 300

Using the algebraic identity

(a + b)^2 = a^2 + b^2 + 2ab

allows determining

ab

directly without solving for individual side lengths.

4

Calculate the area of the right-angled triangle

\text{Area} = 150\text{ cm}^2

The area of a right-angled triangle with perpendicular sides

a

and

b

is given by

\frac{1}{2}ab

.

Key Concept

Perimeter and Area of Right-Angled Triangles using Algebraic Identities

Question 173Question →

A point $P(x, y)$ moves in the Cartesian plane such that it is always equidistant from the two fixed points $A(2, 1)$ and $B(6, 5)$ . If the locus of $P$ intersects the line $2x + y = 14$ at the point $(x_0, y_0)$ , what is the value of $x_0$ ?

Show answer & explanation

Answer: 7

Answer

The value of

x_0

is

7

.

The locus of points equidistant from

A(2, 1)

and

B(6, 5)

is the perpendicular bisector of line segment

AB

. The midpoint of

AB

is

(4, 3)

and its slope is

1

, giving the perpendicular bisector a slope of

-1

. The equation of this locus is

y - 3 = -1(x - 4)

, or

x + y = 7

. Subtracting

x + y = 7

from

2x + y = 14

directly gives

x_0 = 7

.

Step-by-Step Solution

1

Determine the equation of the locus of point P

The locus of P is the perpendicular bisector of segment AB, represented by

x + y = 7

.

The set of all points equidistant from two fixed points forms the perpendicular bisector of the line segment connecting those points.

2

Find the point of intersection with the line

2x + y = 14

Solving

x + y = 7

and

2x + y = 14

simultaneously gives

x_0 = 7

.

The intersection point of two geometric lines must satisfy both equations simultaneously.

Key Concept

Perpendicular Bisector Locus and Line Intersections

Question 174Question →

A cargo ship departs from port $M$ and sails $24\text{ km}$ on a bearing of $050^\circ$ to reach point $N$ . From point $N$ , the ship changes course and sails $10\text{ km}$ on a bearing of $140^\circ$ to reach point $P$ . What is the direct distance, in kilometers, from port $M$ to point $P$ ?

Show answer & explanation

Answer: 26

Answer

The direct distance from port M to point P is 26 km.

The back bearing from N to M is 230°, and the bearing from N to P is 140°. The interior angle at N is 230° - 140° = 90°. Using the Pythagorean theorem for the right triangle formed by M, N, and P, the direct distance is √(24² + 10²) = √676 = 26 km.

Step-by-Step Solution

1

Calculate the interior angle

\angle MNP

at point

N

\angle MNP = (050^\circ + 180^\circ) - 140^\circ = 230^\circ - 140^\circ = 90^\circ

The back bearing from

N

to

M

is

230^\circ

. Subtracting the forward bearing to

P

(

140^\circ

) gives the enclosed interior angle.

2

Apply the Pythagorean theorem to right-angled triangle

MNP

MP = \sqrt{MN^2 + NP^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26\text{ km}

Because

\angle MNP = 90^\circ

, triangle

MNP

is a right-angled triangle with hypotenuse

MP

.

Key Concept

Bearings and Right-Angled Triangles

Question 175Question →

A sector of a circle of radius $21\text{ cm}$ subtends an angle of $60^\circ$ at the centre of the circle. What is the total perimeter of the sector? (Take $\pi = \frac{22}{7}$ )

22\text{ cm}

43\text{ cm}

64\text{ cm}

231\text{ cm}

Show answer & explanation

Answer: $64\text{ cm}$

Answer

The total perimeter of the sector is

64\text{ cm}

.

The perimeter of a sector consists of its curved arc length plus its two bounding radii (

2r

). Using

\theta = 60^\circ

,

r = 21\text{ cm}

, and

\pi = \frac{22}{7}

, the arc length is

\frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22\text{ cm}

. Adding

2 \times 21\text{ cm} = 42\text{ cm}

yields

64\text{ cm}

.

Step-by-Step Solution

1

Calculate the arc length (

s

) of the sector using the formula

s = \frac{\theta}{360^\circ} \times 2\pi r

.

s = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22\text{ cm}

.

The arc length represents the curved boundary of the sector.

2

Calculate the total perimeter of the sector by adding the arc length and the two bounding radii:

P = s + 2r

.

P = 22\text{ cm} + 2(21\text{ cm}) = 22 + 42 = 64\text{ cm}

.

A sector boundary consists of the curved arc plus the two straight radial edges.

Key Concept

Perimeter of a circular sector

Estimated Time:1m 30s

Question 176Question →

Two fixed points in a Cartesian plane are given as $A(-1, 2)$ and $B(3, 4)$ . A point $P(x, y)$ moves in the plane such that $PA^2 + PB^2 = 26$ . Which of the following equations represents the locus of $P$ ?

x^2 + y^2 - 2x - 6y + 2 = 0

x^2 + y^2 + 2x + 6y + 2 = 0

x^2 + y^2 - 4x - 12y + 4 = 0

x^2 + y^2 - 2x - 6y + 30 = 0

Show answer & explanation

Answer: $x^2 + y^2 - 2x - 6y + 2 = 0$

Answer

The equation representing the locus of

P

is

x^2 + y^2 - 2x - 6y + 2 = 0

.

Using the Cartesian coordinate distance formula,

PA^2 = (x+1)^2 + (y-2)^2

and

PB^2 = (x-3)^2 + (y-4)^2

. Adding these together yields

2x^2 + 2y^2 - 4x - 12y + 30

. Setting this equal to 26 gives

2x^2 + 2y^2 - 4x - 12y + 4 = 0

. Dividing the entire equation by 2 yields

x^2 + y^2 - 2x - 6y + 2 = 0

, which is the correct locus equation.

Step-by-Step Solution

1

Express

PA^2

using the distance formula between

P(x, y)

and

A(-1, 2)

.

PA^2 = (x - (-1))^2 + (y - 2)^2 = (x + 1)^2 + (y - 2)^2 = x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 + 2x - 4y + 5

The square of the distance between two points

(x_1, y_1)

and

(x_2, y_2)

is

(x_2 - x_1)^2 + (y_2 - y_1)^2

.

2

Express

PB^2

using the distance formula between

P(x, y)

and

B(3, 4)

.

PB^2 = (x - 3)^2 + (y - 4)^2 = x^2 - 6x + 9 + y^2 - 8y + 16 = x^2 + y^2 - 6x - 8y + 25

Expanding the distance squared formula for point

B

.

3

Sum

PA^2

and

PB^2

and equate to the given constant 26.

(x^2 + y^2 + 2x - 4y + 5) + (x^2 + y^2 - 6x - 8y + 25) = 26 \implies 2x^2 + 2y^2 - 4x - 12y + 30 = 26

Substitute the algebraic expressions into the locus condition

PA^2 + PB^2 = 26

.

4

Rearrange into general circle form and divide through by 2.

2x^2 + 2y^2 - 4x - 12y + 4 = 0 \implies x^2 + y^2 - 2x - 6y + 2 = 0

Subtract 26 from both sides and divide all terms by 2 to obtain the standard Cartesian equation.

Key Concept

Locus of a point with sum of squared distances to two fixed points equal to a constant

Question 177Question →

A hiker starts at camp $C$ and walks $5\text{ km}$ due East to checkpoint $A$ . From checkpoint $A$ , the hiker then walks $3\text{ km}$ on a bearing of $150^\circ$ to reach checkpoint $B$ . What is the direct distance from camp $C$ to checkpoint $B$ ?

7\text{ km}

\sqrt{34}\text{ km}

\sqrt{19}\text{ km}

8\text{ km}

Show answer & explanation

Answer: $7\text{ km}$

Answer

7\text{ km}

The interior angle at checkpoint

A

is calculated using three-figure bearings as

270^\circ - 150^\circ = 120^\circ

. Applying the Cosine Rule

CB^2 = 5^2 + 3^2 - 2(5)(3)\cos(120^\circ)

yields

25 + 9 + 15 = 49

. Taking the square root gives the direct distance of

7\text{ km}

.

Step-by-Step Solution

1

Determine the interior angle

\angle CAB

at checkpoint

A

\angle CAB = 120^\circ

Due East corresponds to a bearing of

090^\circ

. Coming into

A

from

C

means line

AC

points West (

270^\circ

). The bearing of

B

from

A

is

150^\circ

. The interior angle between vector

AC

pointing West (

270^\circ

) and vector

AB

on bearing

150^\circ

is

270^\circ - 150^\circ = 120^\circ

.

2

Apply the Cosine Rule to find length

CB

CB^2 = CA^2 + AB^2 - 2(CA)(AB)\cos(\angle CAB)

We have two sides (

CA = 5\text{ km}

,

AB = 3\text{ km}

) and the included angle (

\angle CAB = 120^\circ

).

3

Substitute the known values into the Cosine Rule formula

CB^2 = 5^2 + 3^2 - 2(5)(3)\cos(120^\circ) = 25 + 9 - 30(-0.5) = 34 + 15 = 49

Since

\cos(120^\circ) = -\cos(60^\circ) = -0.5

, the negative sign inside the cosine product cancels with the subtraction sign in the formula.

4

Take the positive square root to find the distance

CB = \sqrt{49} = 7\text{ km}

Distance must be positive.

Key Concept

Cosine Rule for non-right triangles in bearings problems

Question 178Question →

Two fishing boats leave a harbor $H$ at the same time. Boat $A$ travels on a bearing of $020^\circ$ for a distance of $8\text{ km}$ , while Boat $B$ travels on a bearing of $140^\circ$ for a distance of $7\text{ km}$ . What is the distance between Boat $A$ and Boat $B$ in kilometers?

Show answer & explanation

Answer: 13

Answer

13 km

The distance between the two boats forms the third side of triangle

HAB

, where

HA = 8\text{ km}

,

HB = 7\text{ km}

, and the included angle at the harbor

H

is

\angle AHB = 140^\circ - 020^\circ = 120^\circ

. By the Cosine Rule,

AB^2 = 8^2 + 7^2 - 2(8)(7)\cos(120^\circ) = 64 + 49 - 112(-0.5) = 169

. Taking the square root gives

AB = 13\text{ km}

.

Step-by-Step Solution

1

Calculate the included angle between the direction vectors of the two boats from the harbor.

Included angle

\angle AHB = 140^\circ - 020^\circ = 120^\circ

The angle between two bearings originating from the same point is the difference between their bearing angles.

2

State the Cosine Rule for side

AB

in triangle

HAB

.

AB^2 = HA^2 + HB^2 - 2 \cdot HA \cdot HB \cdot \cos(\angle AHB)

The Cosine Rule calculates an unknown side when two sides and their included angle (SAS) are given.

3

Substitute given side lengths

HA = 8\text{ km}

,

HB = 7\text{ km}

, and angle

\angle AHB = 120^\circ

.

AB^2 = 8^2 + 7^2 - 2(8)(7)\cos(120^\circ) = 64 + 49 - 112(-0.5) = 169

Since

120^\circ

is in the second quadrant,

\cos(120^\circ) = -0.5

, which changes the subtracted term to addition.

4

Compute the principal square root of 169.

AB = \sqrt{169} = 13\text{ km}

Distance is a non-negative scalar quantity.

Key Concept

Applying the Cosine Rule to solve bearing problems

Question 179Question →

What are all the values of $\theta$ in the interval $0^\circ \le \theta \le 360^\circ$ that satisfy the trigonometric equation $4\sin^2\theta - 3 = 0$ ?

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

60^\circ \text{ and } 120^\circ

30^\circ, 150^\circ, 210^\circ, 330^\circ

60^\circ \text{ and } 300^\circ

Show answer & explanation

Answer: $60^\circ, 120^\circ, 240^\circ, 300^\circ$

Answer

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

Solving

4\sin^2\theta - 3 = 0

gives

\sin^2\theta = \frac{3}{4}

, so

\sin\theta = \pm\frac{\sqrt{3}}{2}

. The reference angle for which

\sin\theta = \frac{\sqrt{3}}{2}

is

60^\circ

. The positive root

\sin\theta = +\frac{\sqrt{3}}{2}

gives solutions in Quadrants I and II:

60^\circ

and

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

. The negative root

\sin\theta = -\frac{\sqrt{3}}{2}

gives solutions in Quadrants III and IV:

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

and

360^\circ - 60^\circ = 300^\circ

. Combining these yields all four angles:

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

.

Step-by-Step Solution

1

Isolate the squared trigonometric term in the equation

4\sin^2\theta = 3 \implies \sin^2\theta = \frac{3}{4}

Rearranging the equation allows solving for

\sin\theta

directly.

2

Take the square root of both sides, keeping both positive and negative roots

\sin\theta = \pm\sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2}

Taking the square root of a squared quantity yields both positive and negative values.

3

Find solutions for

\sin\theta = +\frac{\sqrt{3}}{2}

in Quadrants I and II

\theta = 60^\circ \text{ and } \theta = 180^\circ - 60^\circ = 120^\circ

Sine is positive in Quadrants I and II.

4

Find solutions for

\sin\theta = -\frac{\sqrt{3}}{2}

in Quadrants III and IV

\theta = 180^\circ + 60^\circ = 240^\circ \text{ and } \theta = 360^\circ - 60^\circ = 300^\circ

Sine is negative in Quadrants III and IV.

5

Combine all solutions within the domain

0^\circ \le \theta \le 360^\circ

\theta \in \{60^\circ, 120^\circ, 240^\circ, 300^\circ\}

All four angles satisfy the original quadratic trigonometric equation.

Key Concept

Solving quadratic trigonometric equations by finding reference angles and evaluating solutions across all four quadrants

Estimated Time:1m 30s

Question 180Question →

A field surveyor starts at point $P$ and walks $10\text{ km}$ due East to point $Q$ . From point $Q$ , she changes direction and walks $10\text{ km}$ on a bearing of $210^\circ$ to reach point $R$ . What is the bearing of point $P$ from point $R$ ?

330^\circ

060^\circ

150^\circ

300^\circ

Show answer & explanation

Answer: $330^\circ$

Answer

The bearing of point

P

from point

R

is

330^\circ

.

Point

Q

lies

10\text{ km}

East of

P

(bearing

090^\circ

). From

Q

, moving

10\text{ km}

on bearing

210^\circ

forms an interior angle of

60^\circ

with the line

QP

(which has bearing

270^\circ

). Because

PQ = QR = 10\text{ km}

and the included angle is

60^\circ

,

\triangle PQR

is equilateral, making

\angle QRP = 60^\circ

. The back bearing from

R

to

Q

is

030^\circ

. Subtracting the

60^\circ

interior angle from

030^\circ

yields a bearing of

330^\circ

for point

P

from point

R

.

Step-by-Step Solution

1

Determine the interior angle

\angle PQR

at point

Q

.

\angle PQR = 60^\circ

Since

P

is due West of

Q

, the bearing of

P

from

Q

is

270^\circ

. The bearing of

R

from

Q

is

210^\circ

. The interior angle between these two lines is

270^\circ - 210^\circ = 60^\circ

.

2

Analyze the properties of triangle

PQR

.

Triangle

PQR

is an equilateral triangle with side lengths

10\text{ km}

and interior angles of

60^\circ

.

Given

PQ = 10\text{ km}

and

QR = 10\text{ km}

,

\triangle PQR

is isosceles. Since the vertex angle

\angle PQR = 60^\circ

, the remaining two angles are also

60^\circ

each.

3

Calculate the back bearing of

Q

from

R

.

Bearing of

Q

from

R

is

030^\circ

.

The bearing of

R

from

Q

is

210^\circ

. The back bearing is

210^\circ - 180^\circ = 030^\circ

.

4

Compute the bearing of

P

from

R

.

Bearing of

P

from

R

is

330^\circ

.

From line

RQ

(bearing

030^\circ

), line

RP

lies

60^\circ

counter-clockwise (since

\angle QRP = 60^\circ

). Thus,

030^\circ - 60^\circ = -30^\circ \equiv 330^\circ

.

Key Concept

Bearings and Triangle Geometry

Geometry and Trigonometry

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