TSA + Volume
Elevation + Depression
Trig + Bearings
Unit circle
Mixed bag
100

The TSA and volume of:

A rectangular prism with a length of 20cm, a width of 12 cm and a height of 15cm

TSA = 2(lw+wh+hl)

TSA = 1440cm2

V = LWH

V = 3600 cm3

100

Where are angles of elevation and depression always measured from?

The horizontal

100

Convert S40*E to a true bearing

140*T

100

What is the radius of the unit circle?

1

100

TECH FREE

14.6 + 23.08 + 105


142.68

200

The TSA and volume of:

A cylinder with a radius of 21cm and a height of 30cm

TSA = 2 x pi x r2 + 2 x pi x r x h

TSA = 6729.29cm2

V = pi x rx h

V= 41563.27cm3

200

Lucy looks up at a plane flying overhead. This is an example of what type of angle? 

(Elevation or depression)

Angle of elevation

200

Covert 235*T to a compass bearing

S55*W

200

When writing co-ordinates for the unit circle, what order do they go in?

Hint: instead of (x, y) what are the co-ordinates?

(cos, sin)

200

TECH FREE:

a) 3.14 x 7 

b) 3 x 0.12 x 10 

a) 21.98

b) 3.6

300

The TSA and volume of:

A sphere with a radius of 3m

TSA = 4 x pi x r2

TSA = 113.10m2

V = 4/3 x pi x r3

V = 37.7m3

300
From a point (P), on the ground, the angle of elevation to the top of a tree is 50*. If P is 8 metres from the tree, determine the height of the tree correct to 2 decimal places

tan(50*) = h/8

h = 8 x tan(50*)

h = 9.53m

300

A person takes a bearing to a tree of 045*T. They then walk due east. Would this path get them to the tree?

No, they did not walk along the bearing to the tree, they walked directly east, so they would not reach the tree.

300

Which quadrant of the unit circle are each of the following ratios POSITIVE in?

Sine

Cosine

Tangent

Sine = Q1 + Q2

Cosine = Q1 + Q4

Tangent = Q1 + Q3

300

TECH FREE

A triangle has its largest angle of 136* and its smallest angle of 18*. What is the size of its medium angle?

180 - 136 - 18

= 26*

400

Calculate the area and volume of:

A cone with a radius of 8cm and a slant height of 12cm

TSA = pi x r2+ pi x r x l

TSA = 1017.88cm2

V = 1/3 x pi x r2 x h

V = 804.25cm3

400

The angle of depression from a helicopter, at point H, to a swimmer in distress in the water is 60°. If the helicopter is hovering 800m above sea level, determine how far horizontally the swimmer is from the helicopter. Write your answer in metres correct to 2 decimal places.

tan (60*) = O/A

tan(60*)=800/d

d=800/tan(60*)

d=461.88m

400

A boat travels a distance of 5km from P to Q on a bearing of 035*T. How far east of P is Q?

cos(55)=x/5

x = 5 x cos(55)

x = 2.87km

400

What is the co-ordinate for 30* on the unit circle?

cos = A/H

cos(30) = sqrt3/2

sin = O/H

sin(30) = 1/2

Therefore = (sqrt3/2, 1/2)

400

TECH FREE

A building casts a shadow of 9m. The building is 15m high. A tree nearby casts a shadow that is 3m. How tall is the tree?

Scale factor = 9/3 = 3

Tree height 

 15/? = 3

15/3 = ?

?=5. 

The tree is 5m tall


500

Calculate the TSA and volume of a rectangular prism with a length of 2, a width of 3 and a height of 4, stacked on top of a rectangular prism with a length of 5, a width of 6 and a height of 7.

TSA = 2(lw + wh + hl) + 2(lw + wh + hl) - 2(wl)

TSA = 262cm2

V = LWH + LWH

V = 24 + 210 = 234cm3

500

From a point on a cliff, two boats are observed. If the angles of depression are 58* and 32* and the cliff is 46m above sea level, how far apart are the boats?

tan(32)=x/46

x=46 x tan(32*) = 28.74

tan(32)=46/x

x=46/tan(32) = 73.62

73.62 - 28.74 = 44.88m

500

Bob travels on a bearing of 035*T for 10km. From this new point, he then travels due south. If this final position is exactly due east of his original starting point, how far south did Bob travel?

True bearing of 35* = internal angle of 55*

sin(55)=O/H

sin(55)=x/10

x=10 x sin(55)

x = 8.19m

500

What is the co-ordinate for 315* on the unit circle?

315 is an angle of 45 (360 - 315 = 45)

Q4 cos is positive, sin is negative

cos = A/H

cos(45) = 1/sqrt2 = sqrt2/2

sin = O/H

sin(45) = 1/sqrt2 = sqrt2/2

Therefore = (sqrt2/2, - sqrt2/2)

500

TECH FREE

A farmer had a water trough that is the shape of a cylinder cut in half. If the radius of the trough of 1m and the height is 10m, what is the volume of water the farmer can put in there?

Let pi = 3.14

V of half cylinder = 1/2 x pi x r2 x h

V = 1/2 x 3.14 x 12 x 10

V = 1/2 x 31.4

V = 15.7m3

M
e
n
u