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SS1: MATHEMATICS - 3RD TERM

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  1. Geometry (Triangles & Polygons) I
    2 Topics
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    1 Quiz
  2. Geometry (Triangles & Polygon) II
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    1 Quiz
  3. Geometry (Triangles & Polygon) III
    3 Topics
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    1 Quiz
  4. Trigonometry I
    2 Topics
  5. Trigonometry II
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    1 Quiz
  6. Trigonometry III
    3 Topics
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    1 Quiz
  7. Mensuration | Plane Shapes
    3 Topics
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    1 Quiz
  8. Mensuration | Arcs, Sectors and Segments of Circles
    4 Topics
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    1 Quiz
  9. Mensuration | Solid Shapes
    8 Topics
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    1 Quiz
  10. Statistics
    2 Topics
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Topic Content:

  • Special Angles

45°

Consider an Isosceles right-angled triangle of equal sides of 1 unit each and hypotenuse side \( \scriptsize \sqrt {2}\)units.

  • sin 45° = \( \frac {1}{\sqrt {2}}\)
  • cos 45° = \( \frac {1}{\sqrt {2}}\)
  • tan 45° = \( \frac {1}{1} = \scriptsize 1\)

30° and 60°

Consider an equilateral ∆ of side 2 units.

  • sin 30° = \( \frac {1}{2}\)
  • cos 30° = \( \frac {\sqrt {3}}{2}\)
  • tan 30° = \( \frac {1}{\sqrt {3}}\)
  • sin 60° = ...

 

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SOLUTION (a)

Note: We are looking for small letters x and y

Solution:

First, find x

YZ = opposite = x

XZ =  √3 = hypotenuse

θ =  30°

sin 30° = \( \frac {opp}{hyp} = \frac {YZ}{XZ} = \frac {x}{\sqrt {3}}\)

\( \frac {1}{2} =  \frac {x}{\sqrt {3}} \\ \frac {x}{\sqrt {3}} = \frac {1}{2} \\ \)

Multiply both sides by √3

⇒ \( \frac {x}{\sqrt {3}} \scriptsize \; \times \;  \sqrt {3} = \normalsize \frac {1}{2}  \scriptsize \; \times \;  \sqrt {3} \)

⇒ \(\scriptsize \therefore x =  \normalsize \frac{\sqrt {3}} {2} \)

Then find y,

XY = adjacent = y

XZ =  √3 = hypotenuse

θ =  30°

We know the adjacent and hypotenuse so we use cos

cos 30° = \( \frac {adj}{hyp} = \frac {XY}{XZ} \\  \scriptsize cos 30^{\circ} = \normalsize \frac {y}{\sqrt {3}} \)

Remember,

cos 30° = \( \frac {\sqrt {3}}{2}\)

substitute cos 30° into the equation

∴ \( \frac {\sqrt {3}}{2} = \frac {y}{\sqrt {3}} \)

move the unknown y to the left-hand side

∴ \(  \frac {y}{\sqrt {3}} = \frac {\sqrt {3}}{2} \)

Multiply both sides by √3

∴  \( \frac {y}{\sqrt {3}} \scriptsize\; \times \;  \sqrt {3} = \normalsize\frac {\sqrt {3}}{2} \scriptsize\; \times \;  \sqrt {3} \)

∴  \( \scriptsize y = \normalsize\frac {\sqrt {3} \; \times \; \sqrt {3}}{2} \)

⇒ \( \scriptsize y = \normalsize\frac {3}{2} \)

or

\( \scriptsize y = 1\normalsize\frac {1}{2} \)

SOLUTION (b)

Note: We are looking for small letters x and y

To find x we use \( \scriptsize \bar {ZXY} \)

YZ = opposite = x

XY =  6cm = hypotenuse

θ =  30°

sin 30° = \( \frac {opp}{hyp} = \frac {YZ}{XY} = \frac {x}{6}\)

⇒ \( \frac {1}{2} =  \frac {x}{6}\)

move the unknown x to the left-hand side

⇒ \( \frac {x}{6} =  \frac {1}{2}\)

Multiply both sides by 6

⇒ \( \frac {x}{6}  \scriptsize \: \times \: 6 =  \normalsize \frac {1}{2} \scriptsize \: \times \: 6 \)

x = \( \frac {6}{2}\)

∴  x = 3 cm

 

To find y we have to first find XZ then WZ then subtract.

∴ y = XZ – WZ

First, find XZ

To find XZ we use \( \scriptsize \bar {ZXY} \)

XZ =  Adjacent

YZ  = x = 3cm = Opposite

θ =  30°

tan 30° = \( \frac {opp}{adj} =  \frac {YZ}{XZ} \)

tan 30° = \( \frac {x}{XZ}  \)

Remember tan 30° = \( \frac {1}{\sqrt{3}}\)

x = 3cm

⇒ \( \frac {1}{\sqrt{3}} =  \frac {3}{XZ} \)

move the unknown XZ to the left-hand side

⇒ \(  \frac {3}{XZ} = \frac {1}{\sqrt{3}} \)

Cross multiply

⇒ \(  \scriptsize XZ  = 3 \sqrt{3} \: cm\)

 

The next step is to find WZ

To find WZ we use \( \scriptsize \bar {ZWY} \)

Adjacent = WZ

Opposite = x = 3cm (solved previously)

θ =  60°

∴ tan θ = \( \frac {opp}{adj} \)

∴ tan 60° = \( \frac {3}{WZ} \)

Remember tan 60° = \( \frac {\sqrt{3}}{1} \)

∴ \( \frac {\sqrt{3}}{1} = \frac {3}{WZ}  \)

Cross multiply

⇒ \( \scriptsize WZ \sqrt{3} = 3 \)

Divide both sides by √3

⇒ \( \frac{WZ \sqrt{3}}{\sqrt{3}} = \frac{3}{\sqrt{3}}  \)

∴ \( \scriptsize WZ  = \normalsize \frac{3}{\sqrt{3}}  \)

Therefore, y = XZ – WZ

⇒ \(  \scriptsize XZ  = 3 \sqrt{3} \)

⇒ \( \scriptsize WZ = \frac{3}{\sqrt{3}}  \)

y = \(  \scriptsize  3 \sqrt{3} \: – \:  \normalsize \frac{3}{\sqrt{3}} \)

y = \( \frac { 3 \sqrt{3} \: \times \: \sqrt{3} \:   – \: 3}{\sqrt{3}} \)

y = \( \frac { 3  \: \times \:  3 \:  – \: 3}{\sqrt{3}} \)

y = \( \frac { 9 \:  – \: 3}{\sqrt{3}} \)

y = \( \frac {6}{\sqrt{3}} \)

SOLUTION (c)

 

Find small letters x and y

To find x we use the triangle \( \scriptsize \bar {CDB} \)

BC = x = opposite

BD = 4√3 = hypotenuse

CD = adjacent

θ = 60°

sin 60° = \( \frac {opp}{hyp} \\ =\frac {BC}{BD} \\ = \frac {x}{4 \sqrt{3}}\)

Remember sin 60° = \( \frac{\sqrt{3}}{2} \)

∴ \( \frac {\sqrt{3}}{2} =  \frac {x}{4 \sqrt{3}}\)

Cross Multiply

⇒ \( \scriptsize x = \normalsize \frac {4 \sqrt{3} \; \times \sqrt{3} }{2} \)

⇒ \( \scriptsize x = \normalsize \frac {2 \sqrt{3} \; \times \sqrt{3} }{1} \)

i.e. x = 2 × 3

Hence, x = 6cm

 

To find y use the triangle \( \scriptsize \bar {CAB} \)

BA = y = hypotenuse

BC = x = 6cm = opposite

θ = 45°

sin 45° = \( \frac {opp}{hyp} =\frac {BC}{BA} \\= \frac {x}{y}\\ = \frac {6}{y} \)

Remember Sin 45° = \( \frac {1}{\sqrt{2}}\)

⇒ \( \frac {1}{\sqrt{2}} =  \frac {6}{y}\)

Cross Multiply

⇒ \( \scriptsize y = 6 \sqrt {2} cm \)

SOLUTION (Example 2)

The perpendicular distance between the two faces is BD. We therefore have to find the length of DB

We also know DB = 2 DG

or DB = 2 BG

or DB = DG + BG

This is because DG and BG have the same value

The first step is to find DG which is \( \frac{1}{2}\) of BD

sin 60° = \( \frac {opp}{hyp} = \frac {DG}{DC} = \frac {DG}{8}\)

sin 60° = \( \frac {\sqrt{3}}{2} \)

∴ \( \frac {\sqrt{3}}{2} =  \frac {DG}{8}\)

⇒ \( \scriptsize DG = \normalsize \frac {8 \sqrt{3} }{2} \)

⇒ \( \scriptsize DG = 4 \sqrt{3} \)

DB = 2DG

= \( \scriptsize 2 \: \times \: 4  \sqrt{3} \)

DB = \( \scriptsize 8  \sqrt{3} \)

SOLUTION (Example 3)

Question

From a place 400 m north of X, a student walks eastwards to a place Y which is 800 m from X. What is the bearing of X from Y? (WAEC)

Solution

Let’s call the place where the students start to walk from Z. Z is north of X so XZ is a straight line upwards. XZ = 400 m

He then walks Eastward to place Y. Y is 800 m from X so we connect the line between X and Y and write 800 m next to it.

 

We know XZ = 400 m and XY is 800 m. The question is asking for the bearing of X from Y which is the angle shown in the diagram below. You can see that we named two angles  \( \scriptsize \bar {YXZ}  \) because they are alternate angles and are equal.

So first, we calculate \( \scriptsize \bar {YXZ}  \), and then we will be able to calculate the bearing of X from Y.

So from the diagram below, we can calculate \( \scriptsize \bar {YXZ}  \)

⇒ \( \scriptsize \bar {YXZ} = \theta \)

Adjacent = 400 m

Hypotenuse = 800 m

⇒ \(\scriptsize cos \theta =  \normalsize \frac {adj}{hyp} \\ = \frac {400}{800}\)

cos θ = (0.5000) = \( \frac{1}{2} \)

but cos 60° = \( \frac{1}{2} \)

∴ θ = 60°

⇒ \( \scriptsize \bar {YXZ} = 60^{/circ} \)

 

We can then redraw the diagram as shown below.

The bearing of X from Y = (90° + 90° + θ)

Y = (90° + 90° + 60°)

= 180° + 60°

= 240°

SOLUTION (Example 4)

Question

If tan θ = \( \frac{8}{15} \),    find the value of \( \frac{sin \theta \: +\: cos \theta}{Cos \theta(1\: – \: Cos \theta)} \)

Solution

If tan θ = \( \frac{8}{15} \)

We can draw a right-angle triangle from this information because

tan θ = \( \frac{opp}{adj} \)

8 = opposite, 15 = adjacent

Using Pythagoras Theorem

⇒ \( \scriptsize AB = \sqrt {8^2 + 15^2} \)

⇒ \( \scriptsize AB = \sqrt {289} = 17cm \)

considering the angle \( \scriptsize \bar{CAB} = \theta \)

adjacent =  AC = 15

hypotenuse = AB =  17

opposite =  CB = 8

cos θ° = \( \frac {adj}{hyp}\\ = \frac {AC}{AB}\\ = \frac {15}{17}\)

sin θ° = \( \frac {opp}{hyp} \\ = \frac {CB}{AB}\\ = \frac {8}{17}\)

 

Substitute these values into the given equation

⇒ \( \frac{sin \theta \: +\: cos \theta}{Cos \theta(1\: – \: Cos \theta)} \)

⇒ \( \frac{sin \theta \: +  \: cos\theta }{Cos \theta (1\: -\: Cos \theta)} \\ =  \large \frac {\frac {8}{17} \: + \:  \frac {15}{17}}{\frac {15}{17} \left ( \scriptsize 1 \: –  \: \normalsize \frac {15}{17}  \right )}  \)

= \( \large \frac { \frac{23}{17}}{\frac {15}{17} \left (  \frac {17 \: -\: 15}{17}  \right )}  \)

= \( \large \frac { \frac{23}{17}}{\frac {15}{17} \left (  \frac {2}{17}  \right )}  \)

= \( \large \frac { \frac{23}{17}}{\frac {30}{289}}  \)

= \(  \frac{23}{17}  \:  \times \:  \frac {289}{30} \)

= \(  \frac{23}{1} \:  \times \: \frac {17}{30} \)

= \(  \frac{391}{30} \scriptsize \:  0r \: 13 \normalsize \frac {1}{30} \)

SOLUTION (Example 5)

Question

A vertical pole, XY is erected on a piece of level ground. A student whose eye is 1.5 m above the ground is standing at Z, 12 m away from Y, the foot of the pole. If XEY = 50º, calculate the height of the top of the pole above the ground level. Give your answer correct to the nearest metre.

Solution

 

The height of the Pole = XY

eye is 1.5 m from ground

Study and make sure you understand the diagram.

mY + mX = XY

mY = 1.5 m

mX = k = ?

We need to find k in order to find XY

To find K we can use the angle  \(\scriptsize \bar{XEm} \)which is not given. The angle given in the question is \( \scriptsize \bar{XEY}  = 50^{/circ}\)

To find \(\scriptsize \bar{XEm} \) we need to first find \(\scriptsize \bar{mEY} \) then subtract this value from 50º

Triangle \(\scriptsize \bar{mEY} \)

θ = \(\scriptsize \bar{mEY} \)

opposite = mY = 1.5m

adjacent = mE = 12 cm

tan θ = \( \frac{opp}{adj} \\ = \frac{1.5}{12}\\ = \scriptsize 0.125 \)

tan θ = (0.125)

tan -1  θ  = 7.125016

θ = 7°

⇒ \(\scriptsize \bar{mEY} \) =  θ = 7°

⇒ \( \scriptsize X\bar{E}Y = X\bar{E}m \: + \: m \bar{E}Y \)

50° = XEm + 7°

XEm = 50 – 7

= 43°

We can then find k, now that we know \(\scriptsize \bar{XEm} \)

Triangle \(\scriptsize \bar{XEm} \)

opposite = mX = k

adjacent = mE = 12m

XEm = θ = 43°

tan 43° = \( \frac{opp}{adj} \\ = \frac{k}{12}  \)

⇒ \( \frac{k}{12} = \scriptsize tan 43^o  \)

k = 12 × tan 43º

=12 × 0.9325

k = 11.1901

Height of pole = XY

XY = mY + mX

XY =  1.5 + k

XY  = 1.5 + 11.11901

XY  = 12.69

XY  =12.7

Answer: XY  =13 m (nearest metre)

SOLUTION (Example 6)

Question

The shadow of a post is 6m longer when the elevation of the Sun is 30° than when it is 60°. Calculate the height of the post.

Solution

Consider  \( \scriptsize \Delta A\bar{C} B\)

opposite = AB = h 

Adjacent BC = x

θ = 60°

tan 60° = \( \frac{opp}{adj}  \)

tan 60° = \( \frac{h}{x}  \)

⇒ \( \frac{h}{x}  \) = tan 60°

cross multiply

i.e. h = x tan 60°

i.e. h = x √3 ……….. (i)

also consider \( \scriptsize \Delta A\bar{D} B\)

opposite = AB =  h

adjacent = BD = x + 6

θ = 30°

tan 30° = \( \frac{opp}{adj} \\ = \frac{h}{x + 6}  \)

tan 30° = \( \frac{1}{\sqrt{3}}  \)

∴ \( \frac{1}{\sqrt{3}} = \frac{h}{x + 6}   \)

cross multiply

x + 6 = h√3  ………. (ii)

Substitute for h (equation (i)) in equation (ii)

i.e. x + 6 =  x√3  × √3

x + 6 = 3x

i.e. 2x = 6

i.e. x = 3m

Substitute for x in equation (i)

i.e. h = x√3 , 

Answer:  h = 3√3 m

SOLUTION (Example 7)

Question

A man sets out to travel from A to C via B. From A he travels a distance of 8 km on a bearing N30ºE to B. From B he travels a further 6 km due east.

Solution

Special angle

(i) Calculate how far C is north of A

How far C is north of A is the line DA shown in the diagram above.

Using the triangle BAD in purple we can calculate DA

hypotenuse = 8km

adjacent = DA

θ = 30º

From the values, we can use cos

Cos θ = \( \frac {adj}{hyp} \)

Cos 30º = \( \frac {DA}{8km} \)

:> \( \frac {DA}{8km} \) = Cos 30º

DA = \( \scriptsize 8 \times  \normalsize \frac{\sqrt{3}}{2} \)

DA = 4√3

Answer: C is 4√3 km north of A

 

(ii) Calculate how far C is East of A.

How far C is east of A is the line CD shown in the diagram above.

CB + BD = CD

CB is given = 6 km, therefore we need to find BD

Using the triangle BAD in purple we can calculate BD

hypotenuse = 8 km

opposite = BD

θ = 30º

From the values, we can use sin

sin 30° = \( \frac{opp}{hyp} = \frac{BD}{8} \)

⇒ \( \frac{BD}{8} \) = sin 30°

BD = \( \scriptsize 8 \times  \normalsize \frac{1}{2} \)

BD = 4

CD = CB + BD

CB = 6km, BD = 4km

∴ CD = 6km + 4km

CD = 10km

 

 

2. Hence, or otherwise, calculate the distance AC correct to 1 decimal place.

Using Pythagoras Theorem

⇒ \(\scriptsize \bar{AC}^2 =  CD^2 + DA^2\)

CD = 6 km + 4 km = 10 km

DA = 4√3 km

AC = ?

⇒ \(\scriptsize \bar{AC} = \sqrt { 10^2 + (4 \sqrt {3})^2}\)

= \( \scriptsize \sqrt {100 + 48} = \sqrt {148} \)

= \( \scriptsize 2 \sqrt {37} = 12.1655 \)

Answer: \( \scriptsize \bar{AC} = 12.2km\)

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