Mathematics

Key unit competence
By the end of the unit, learners should be able to find the length, sides and angles in
right-angled triangles using trigonometric ratios.

Unit Outline

• Definition of a right-angled triangle.

• Elements of a right-angled triangle.

• Relationship between the elements of a right-angled triangle through the
   use of Pythagoras theorem.

• Median and perpendicular heights.

• Determining the sides of right-angled  triangle given their orthogonal
   projections on the hypotenuse.

• Solving problems in right-angled triangles using the element properties
   and the Pythagoras theorem

• Trigonometric ratios in a right-angled triangle: sine, cosine and tangent.

Introduction

Unit Focus Activity

8.1 Review of Pythagoras theorem

    Activity 8.1

1. Draw and label all features and sides of a right angled triangle.

2. Show the relationship between the three sides of a right angled triangle.
  

In a right-angled triangle, one angle is 90°. The longer side in a right-angled triangle
 called the hypotenuse and the two shorter sides (legs).
Consider the right-angled triangle in Fig 8.3 below.

Pythagoras theorem states that the square of the hypotenuse (the side
opposite the right angle) is equal to the sum of the squares of the other two sides.
Pythagoras theorem

a² + b² = c²

Solution

(a) Triangle (a)

Use Pythagoras theorem
a² + b² = c²
Let a = 6 cm, b = 8 cm; c = x cm
6² + 8² = x²
36 + 64 = x²
100 = x2
Finding the square root of 100
x = √100 cm
x = 10 cm

Exercise 8.1
1. Determine, to two decimal places, the length of the third side of the
   right angled triangle where the one side and the hypotenuse have given
   below.
(a) 5 cm, 12 cm

(b) 1 cm, 2 cm

(c) 3 cm, 4 cm

(d) 2.5cm, 3 cm

(e) 1 cm, 1.73 cm

(f) 2 cm, 7 cm

2. The diagram in Fig 8.5 shows a wooden frame that is to be part of the
     roof of a house.

(a) Use Pythagoras theorem in triangle PQR to find the length
   of PQ

(b) Calculate the length QS

3. Fig 8.6 below shows an isosceles triangle with a base of length 4 m and
   perpendicular height 8 m. Calculate length labeled x of one of the equal
   sides.

4. The diagram in Fig 8.7 shows a vertical flagpole of height 5.2 m with
   a rope tied to the top. When the rope is pulled tight, the bottom end is
   3.8 m from the base of the flagpole. Calculate the length of the rope.

8.2 Median theorem of a rightangled triangle

Activity 8.2

1. Using a ruler and pair of compass or protractor, draw any rectangle
    ABCD with measurements of your choice.

2. Draw the diagonals AC and BD and label the point of their intersection
    as E.

3. Measure and compare the diagonal segments BE and ED, and AE with
    EC. What do you notice? What can you say about the diagonals of a
    rectangle?

4. Consider the right angled-triangle ABD in Fig. 8.8 below isolated
    from rectangle ABCD.

(a) Measure and compare the segment BE and DE of the
     hypotenuse. What do you notice?

(b) In what proportions does point
     E divide the hypotenuse DB? What is the name of point E in
     relation to side DB?

(c) What is the name of line AE that is drawn to from vertex A
     to point E?

(d) Measure and compare the length of AE to the hypotenuse
      DB. What do you notice?

Activity 8.3

1. Draw any right-angled triangle PQR, with dimensions of your
   choice and ∠Q = 90°.

2. Measure and locate the midpoint of the hypotenuse PR and label it
    S. Join vertex Q to points with a straight line.

3. Measure and compare the lengths QS and the hypotenuse PR. What
     do you notice?

4. Measure and compare the lengths QS with PS and RS. What do you
    notice?.

Consider a right-angled triangle. A line drawn from any vertex of the triangle to
 the midpoint of the side opposite to that vertex is called a median, Fig. 8.9 below.

XR, ZV and YU are the medians of triangle XYZ (Fig. 8.9).
Let us consider the median XW from the right-angled vertex X to the midpoint W of
the hypotenuse YZ as shown in Fig. 8.10.

We observe the following:
WX = 1/2YZ
Hence WX = WZ = WY.

Theorem
The median theorem of a right-angled
triangle states that: the median from the right angled vertex to the
hypotenuse is half the length of the hypotenuse.

Median = 1/2 Hypotenuse
As such, that median subdivides the
right-angled triangle into two similar
isosceles triangles.

Example 8.2
In a right-angled triangle ABC, line BC is the hypotenuse and AN is 5 cm long.
What is the length of the hypotenuse?

Example 8.3
In a right-angled triangle, the median to the hypotenuse has a length of (3x –
7) cm. the hypotenuse is (5x – 4) long. Find the value x, hence find the length
of the hypotenuse.

Exercise 8.2

1. In a right-angled triangle, the median to the hypotenuse is 4.5 cm. What is
    the length of the hypotenuse?

2. One leg of a right triangle is 12 cm. The median to the hypotenuse is
    7.5 cm. Find the:

(a) length of the hypotenuse.

(b) length of the other leg of the hypotenuse.

3. The two legs of a right-angled triangle  are 4.5 cm and 6 cm long. Find the
    length of the median from the rightangled vertex to the hypotenuse.

4. Triangle KLM is right-angled at vertex L and ∠LKM = 24°. N is the
    midpoint of the hypotenuse KM .  Find the value of angle:

(a) KLN              (b) LNM

5. In a right-angled triangle EFG, the hypotenuse is (3x + 8)cm long. The
    median to the hypotenuse is (5x – 10) long. Find the value of x hence find
   the length of the median.

8.3 Altitude (Height) theorems of a right-angled triangle

Activity 8.4

Materials: Manila paper, protractor, ruler, pencil, a pair of compasses.

1. Draw a right-angled triangle ABC with dimensions of your
    choice on a manila paper.

2. Drop a perpendicular from the  right-angled vertex C to intersect
    the hypotenuse at point N. Line NC is known as the altitude of
    the triangle.

3. Cut off triangles ANC and NBC from the manila paper.

4. Rotate the triangular cut out NBC by 90° anticlockwise.

6. Test triangles ANC and CNB for similarity using equality of
    corresponding angles. What do you notice?

  Activity 8.5

1. Measure the length of the altitude CN and each of the two segments
    of the hypotenuse i.e. AN and NB.

2. Determine and compare the ratios of the corresponding sides
    NC/AN and NB/NC . What do you notice?

A line drawn from any vertex of a triangle to intersect the side opposite
to the vertex perpendicularly is known as an altitude of the triangle.
The altitude from the right-angled vertex of a right-angled triangle to
the hypotenuse divides the triangle into two similar triangles. The
corresponding those two angles are equal.

For example, triangle EFG(Fig 8.15) forms two similar triangle, EHG and
GHF, when cut along the altitude GH as shown in Fig. 8.15.

Consider a right-angled triangle EFG with the altitude drawn from the right-angled
vertex G to the hypotenuse.

This is the mathematical representation of the
altitude theorem of a right-angled triangle.

Thus, the altitude theorem of a rightangled
triangle states that,

“The altitude to the hypotenuse of a right-angled triangle is the mean
proportional between the segments into which it divides the hypotenuse.”

Example 8.5

In a right-angled triangle ABC, AD is the altitude from vertex A to the
hypotenuse. If AD = 6 cm and DC = 9 cm, find the length of segment BD

Exercise 8.3

1. Find the length AD in triangle ABC in Fig. 8.19 given that DC = 10 cm
    and DB = 8 cm.

2. Find the length MP in triangle MNO shown in Fig. 8.20 given that OP = 4
    cm and NP = 12 cm.

3. In the right-angled triangle EFG shown in Fig. 8.21, EH = 4 cm and
    HF = 9 cm. Find the altitude to the hypotenuse.

4. In triangle PQR shown in Fig. 8.22, PQ = 7 cm, PS = 3.6 cm and SQ is
   the altitude to the hypotenuse. Find the length: (a) SQ (b) SR

5. The altitude to the hypotenuse of a right-angled triangle is 8 cm long. If
     the hypotenuse is 20 cm long, what are the lengths of the two segments of
      the hypotenuse?

6. The altitude to the hypotenuse of a right-angled triangle divides the
     hypotenuse into segments that are 12 cm and 15 cm long. Find the
     length of the altitude.

7. Find the value of x in triangle KLM shown in Fig 8.23 given that
    KM = (x + 5) cm, NL = x cm and NM = 6 cm.

8. Triangle EFG is right angled at F. FH is the altitude to the hypotenuse and
   has length of 10 cm. If the lengths of the two segments EH and HG of the
   hypotenuse are in the ratio 1 : 4, find their actual lengths in centimetres.
  (Hint: Sketch the triangle and let EH = x)

9. The length of the hypotenuse of a right-angled triangle is 29 cm.
    If the length of the altitude from the hypotenuse is 10 cm long, find the
   lengths of the two segments of the hypotenuse.

10. The legs of a right-angled triangle are 4.5 cm and 6 cm long. Calculate
     the length of the altitude to the hypotenuse of the triangle.

8.4 Leg theorem of a right-angled triangle

Activity 8.6

1. Draw and cut out on a manilapaper two identical right angled
    triangular cut outs ABC with the dimensions AC = 8 cm, CB = 6
   cm AB = 10 cm and ∠ACB =90°.

2. One of the cut out, drop a perpendicular from vetex C to
    meet the hypotenuse at point D.

3. Cut out the two triangles ADC and DBC and compare each of
    them with the ‘mother’ triangular cut out ABC (the represented by
    the unsubdivided cut out).

4. Test each of similarity with the ‘mother’ triangle using the
    corresponding angles. What do you notice?
    Measure and compare the ratios of:

What do you notice?
Measure and compare the ratios;

What do you notice?

Activity 8.7

Look at the right-angled triangles EFG and PQR in Fig 8.24 and Fig.8.25.

Consider triangle UVW

These observations are summarized into what is known as the leg theorem.
It states that “the leg of a right-angled triangle is the mean proportional between

the hypotenuse and the projection of the leg on the hypotenuse.” This is
mathematically represented as;

Example 8.7

Fig. 8.27 below shows a right angled triangle in which AB = 12 cm, AD = 3 cm
and CD is the attitude to the hypotenuse. Find the length AC.

Example 8.8

Fig. 8.28 shows a right-angled triangle
PQ in which PQ = 6 cm. QT = 3.6 cm.
Find the length of RT. Hence or otherwisefind the length of QR.

Exercise 8.4

1. Find the length of AC in the rightangled
triangle ABC shown in Fig.8.29.

2. What is the length PR in the rightangled triangle PQR shown in Fig.

8.30 below given that PS = 6 cm and PQ = 14 cm?

3. Find the length NP in triangle MNO shown in Fig. 8.31 below. NO = 10 cm.

4. What are the lengths of BD and AC in triangle ABC, Fig. 8.32, given that
   AD = 8 cm and CD = 12 cm?

5. Find the length of WX in triangle  WXY, Fig. 8.33, given that WY = 7 cm
    and YB = 9 cm.

6. The hypotenuse of a right-angled triangle is 5 cm long and the longer
    leg of the triangle is 4 cm long. What is the length of the projection of the
    shorter leg on the hypotenuse?

7. In triangle UVW in Fig. 8.34,
   UX = 6 cm and VW = 8 cm.

Find: (a) XV (b) XW

8. Fig. 8.35 below shows triangle KLM.
Given that KM = 12 cm, ML = 9cm and KL = 15 cm, find;
 (a) KN     (b) NL      (c) NM

9. In Fig. 8.36, O is the centre of the circle. PR = 12 cm and PT = TU = UR.

Find; (a) QR (b) PQ (c) SU
10. The altitude to the hypotenuse of a right-angled triangle EFG divides
     the hypotenuse into 9 cm and 12 cm segments.
  Find the lengths of the:
(a) altitude to the hypotenuse

(b) shorter leg of triangle EFG

(c) longer leg of triangle EFG.

8.5 Introduction to trigonometry

Activity 8.8

With reference to angle θ, name the adjacent side, the opposite side and
the hypotenuse in each of the triangles in Fig. 8.37.

The word ‘trigonometry’ is derived from
two Greek word: trigonon meaning a triangle and metron, meaning measurement. Thus
trigonometry is the branch of mathematics concerned with the relationships between
the sides and the angles of triangles. In trigonometry, Greek letters are used
to indicate, in a general way, the sizes of various angles.
The most commonly used of these letters
are:
α = alpha;     β = beta;            γ = gamma;
δ = delta;      θ = theta;           φ = phi;
ω = omega.

In senior 2, we studied the relationship between the lengths of the sides of a
right-angled triangle, which is known as the Pythagorean relationship. In this unit,
we shall study the relationships between the acute angles and the sides of a rightangled
triangle.

The sides of such a triangle are named, with reference to specific angle e.g. θ, as
shown in Fig. 8.40.

If C is used as the reference angle in Fig 8.41 AB becomes the opposite
side and BC the adjacent side. These
definitions of the sides apply to any rightangled
triangle in any position.

Solution
(a) hypotenuse is QR

(b) (i) PQ    (ii) PR    (iii) PR
    (iv)PQ

Exercise 8.5

8.6 Trigonometric Ratios

In any right-angled triangle, there are three basic ratios between the sides of
the triangle with reference to a particular acute angle in the triangle. They are sine,
cosine and tangent.

8.6.1 Sine and cosine of an acute angle

Activity 8.9

Exercise 8.6

1. By drawing and measuring, find the approximate values for:
(a) sin 20°, cos 20°

(b) sin 42°, cos 42°

(c) sin 65°, cos 65°

(d) sin 78°, cos 78°

2. Find, by drawing and measuring, approximate sizes of the angles whose
sines and cosines are given below.

sin H = 0.84

(i) sin I = 0.65

(j) cos J = 0.23

(k) cos K = 0.34

(l) cos L = 0.56
Hint for parts (g) to (l): First convert the decimals into fractions.

8.6.2.1 Finding sine and cosine using calculators

Activity 8.10

Use your scientific calculator to find the sines and cosines of the following.
(a) 32º   (b) 17.89º   (c) 73.5º
Note that as the angles increase from
0° to 90°,
(i) their sines increase from 0 to 1, and

(ii) their cosines decrease from 1 to 0.

Example 8.11

Use calculators to find the value of:
(a) sin 53.4°   (b) cos 71.2°

Solution

(a) Press sin, type 53.4º press =
(0.802817… is displayed)
Thus, sin 53.4° = 0.802 8

(b) Press cos, type 71.2º press =
(0.3222656… is displayed)
Thus,cos 71.2° = 0.322 3

Example 8.12

Use a calculator to find the angle whose
(a) sine is 0.866

(b) cosine is 0.7071

Solution

(a) Let θ be the angle whose sine is 0.866.
    So, Sin θ = 0.866.
We use sin^-1 function called Sine
inverse to find the value of angle θ
∴ θ = sin-1 (0.866) = 60º.

On your scientific calculator, press shift,
press sine button, type 0.866, press equal
signs.

We get θ = 60°
(b) Let ∝ be the angle whose cosine is
0.7071
So, Cos ∝ = 0.7071
We use cos-1 function called Cosine
inverse to find the value of angle ∝
∴ θ = cos-1 (0.7071) = 45º

On your scientific calculator, press shift, press cosine button, type 0.7071, press
equal signs.

We get θ = 45°

Exercise 8.7
1. Use calculators to find the sine of:
(a) 3°         (b) 13°          (c) 70°

(d) 63°       (e) 13.2°       (f) 47.8°

(g) 79.2°   89.2°

2. Use calculators to find the cosine of:
(a) 18° (b) 27° (c) 49°
(d) 70° (e) 19.5° (f) 36.6°
(g) 77.7° 83.9°
3. Use calculators to find the angle whose sine is:
(a) 0.397 1      (b) 0.788 0

(c) 0.927 8        (d) 0.996 3

(e) 0.948 9       (f) 0.917 8

4. Use calculators to find the angle whose cosine is:
(a) 0.918 2    (b) 0.564 1

(c) 0.123 4     (d) 0.432 1

(e) 0.880 1    (f) 0.555 5

8.6.2.2 Using sines and cosines to find angles and lengths of sides of
    right-angled triangles

Activity 8.11

Use your knowledge of sines and cosines to find the values of unknown
sides and angles in the following triangles.

Exercise 8.8

1. In Fig.8.56, find the lengths marked with letters.The lengths are in
   centimetres. Give your answer correct to 3 decimal places.

2. Find the lengths marked with letters in Fig. 8.57. Measurements are in
    centimetres. Give your answers to 3 significant figures.

8.6.2 Tangent of an acute angle

8.6.2.1 Definition of tangent of an angle

 Activity 8.12

Example 8.16
By drawing and measuring, find the
angle whose tangent is 2/3 .

Solution

Let the angle be θ.
tan θ = opposite/adjacent = 2/3

Exercise 8.9

1. Find the tangents of the following angles by drawing and measuring.
(a) 40°        (b) 45°        (c) 50°

(d) 25°       (e) 33°        (f) 60°
2. By drawing and measuring, find the angles whose tangents are as follows:
(a) 1/4      (b) 2/5        (c) 3/7

(d) 5/4      (e) 6/5         (f) 7/4

(g) 8/3      1             (i) 3
8.6.2.2 Use calculators to find tangent of angles and angles
           given their tangents

Activity 8.13

Use scientific calculator to find tangent of:
(a) 7°     (b) 13.6°       (c) 50°

In lower levels, we learnt how to use calculators in simple mathematical
operations. We can also use calculators to find tangents, cosines or sines of given
angles. The following examples will show how to find tangents of given angles.

Example 8.17

Use calculators to find the tangents of
the following:
(a) 67°    (b) 60.55°    (c) 38.88º.

Solution

(a) (Press tan, type 67 and press) =
      tan 67° = 2.355 9

(b) To find tan of 60.55°:
Press tan enter 60.55
Press =, (1.7711 is displayed)

∴ tan 60.55º = 1.771 1
(c) tan 38.88°
Press tan, type 38.88
Press = (0.8063 is displayed

∴ tan 38.88° = 0.806 3

Example 8.18

Use a calculator to find the angle whose
tangent is:
(a) 1.28 (b) 0.875

Solution

We use the function tan^-1 read as inverse
to find the angle whose tangent is given.

(a) tan^-1 1.28
  Press tan-1, type 1.28 then press =
  We get tan-1 1.28= 52º
  The angle is 52º.

(b) tan-1 0.875
     Press tan-1, type 0.875 then press =
     We get tan-1 0.875= 41.1º
     The angle is 41.1º.

Exercise 8.10

Use a calculator to find the tangents of the
following angles:
(a) 7°            (b) 13.6°       (c) 50°

(d) 47.7°        (e) 80.5°      (f) 2.21°

(g) 18.46°       55.68°    (i) 66.99°

(j) 51°               (k) 64°         (l) 73°

8.6.2.3 Using tangents to find  lengths of sides and angles of
     right-angled triangles

Activity 8.14

Exercise 8.11

1. Find the length of the side indicated in each of the triangles in Fig. 8.66.
     State your answer to 3 s.f.

2. Find the length of the side indicated in each of the triangles in Fig 8.67.
    State your answer in 3 s.f.

3. For this question, use ΔABC which is right angled at B to calculate the
     required angle given the following information;
(a) AB = 8 cm, BC = 4.25 cm.
     Calculate angle BAC.

(b) AB = 12cm, BC = 5 cm.
     Calculate ∠A.

(c) AC = 6 cm, BC = 2.82 cm.
      Calculate ∠A.

(d) AC = 9 cm, AB = 5.03 cm.
     Calculate ∠BAC.

(e) AC = 15 cm BC = 11 cm,
     calculate ∠ACB.

8.6.3 Application of trigonometic ratios (sine, cosine and tangent)

Activity 8.15

Let A be the foot of the tower and x be the height of the tower which can be
viewed from points B and C at angles α and β as Fig 8.68 below shows. Points
B and C are a cm apart.

The following examples illustrate some of the calculations involved when faced with
 real life situations.

Note that 65° is the complement of 25°, and that it is easier to multiply 15 by tan
65° (Method 2) than to divide 15 by tan 25° (Method 1). Hence, given a problem
like the one above, it is better to find the complement of the given angle and then
use it to solve the problem. However, if you are using a calculator, either method will do.

Exercise 8.12

1. A ladder of length 5.5 m rests against a vertical wall so that the angle
   between the ladder and the ground is 60°. How far from the wall is the foot
   of the ladder?

2. A boy is flying a kite using a string of length 56 m. If the string is taut
    and it makes an angle of 62° with the horizontal, how high is the kite?
    (Ignore the height of the boy).

3. Find the dimensions of the floor of a rectangular hall given that the angle
    between a diagonal and the longer side is 25° and that the length of the
    diagonal is 10 m.

4. A plane takes off from an airport and after a while, an observer at the top of

the control tower sees it at an angle of elevation of 9°.
At that instant, the pilot reports that he has attained an altitude of 2.4 km.
If the height of the control tower is 50m, find the horizontal distance that
the plane has flown?

5. After walking 100 m up a sloping road, a man finds that he has risen 30
    m. What is the angle of slope of theroad?

6. The tops of two vertical poles of heights 20 m and 15 m are joined by
    a taut wire 12 m long. What is the angle of slope of the wire?

7. A man walks 1 000 m on a bearing of 025° and then 800 m on a bearing of
    035°. How far north is he from the starting point?

8. Two boats A and B left a holiday resort at the coast. Boat A travelled 4
    km on a bearing of 030° and boat B travelled 6 km on a bearing of 130°.

(a) Find which boat travelled further eastwards and by how much.

(b) How far to the north is boat A from boat B?

9. A bridge crosses a river at an angle of 60°. If the length of the bridge is
   170 m, what is the width of the river?

10. A man sitting at a window with his eye 20 m above the ground just sees
      the sun over the top of a roof 45 m high. If that roof is 30 m away from
      him horizontally, find the angle of elevation of the sun.

11. The shaft of a mine descends for 100 m at an angle of 13° to the horizontal
      and then for 200 m at an angle of 7° to the horizontal. How far below the
      starting point is the end of the shaft?

12. Two girls, one east and the other west of a tower, measure the angles
     of elevation of the top its spire as 28° and 37°. If the top of the spire is 120 m
     high, how far apart are the girls?

13. Jacob and Bernard stand on one side of a tower and in a straight line with
     the tower. They each use a clinometer and determine the angle of elevation
    of the top the tower as 30° and 60° respectively. If their distance apart is
    100 m, find the height of the tower.

14. Fig. 8.71 shows the side view of an ironing board. The legs are all 95 cm
      long and make 60° with the floor when completely stretched.

(a) How high is the ironing surface from the floor, given that the
     board is 2.5 cm thick?

(b) How far apart are the legs at the floor?

Unit Summary

• Pythagoras theorem Consider the right-angled triangle
  ABC shown in Fig 8.72.

Pythagoras theorem states that a2 + b2 = c2

• The median theorem of a rightangled triangle states that.
  The median from the right-angled vertex to the hypotenuse is half the
   length of the hypotenuse. Consider the right angled triangle
    XYZ Fig. 8.73.

• The altitude theorem of a right angled triangle states that:
   “The altitude to the hypotenuse of a right-angled triangle is the mean
    proportional between the segments into which it divides the hypotenuse.”
   Consider the right-angled triangle EFG. Fig. 8.74.

• The leg theorem of a right-angled triangle states that “the leg of a
   right-angled triangle is the mean proportional between the hypotenuse
   and the projection of the leg on the hypotenuse.” It can be simply
   presented as;

• Naming of sides and angles in a right angled triangle

                                          c

Unit 8 Test

1. Find the missing sides in Fig. 8.80. (Measurements are in cm).

2. A ladder which is 6 m long leans against a wall. If the top of the ladder
    is 4 m above the ground, how far from the wall is the foot of the ladder?

3. In a right-angled triangle, the median to the hypotenuse is 6.4 cm. What is
    the length of the hypotenuse.

4. In a right-angled triangle the length of the median to the hypotenuse is

    (3x – 7) cm long. The hypotenuse is
    (5x – 4) cm long. Find the length of the hypotenuse.

5. Find the lengths of the sides marked with letters in the following triangle.

6. A right-angled triangle ABC has its leg a = 5 cm long and altitude to the
    hypotenuse h = 3 cm. Find the length of sides b and c.

7. In a right-angled triangle the altitude
    to the hypotenuse is 8 cm high. The hypotenuse is 20 cm long. Find
    the lengths of the segements of the hypotenuse (Hint: let the length of
    the segment be x.)

8. Find the length AB in the triangle ABC below Fig. 8.82.

9. A boat is 300 m from a vertical cliff. The angle of elevation of the top of
   the cliff is 30°. After the boat moves a distance x metres towards the cliff,
   the angle of elevation becomes 70°. Find the value of x, to the nearest 1 m.

10. Two tall buildings A and B are 40 m apart. From foot A, the angle of
      elevation of the top of B is 60°. From the top of A, the angle of depression
      of the top of B is 30°. Find the heights of A and B, to the nearest 1 m.

11. Find the value of the side marked with letters in the following figures: