The Trigonometric Ratio 1 Questions and Answers

The Trigonometric Ratio 1 Questions

1.  Given sin(90 – a) = ½ , find without using trigonometric tables the value of cos a (2mks)

2.  If ,find without using tables or calculator, the value of

(3 marks)

3. At point A, David observed the top of a tall building at an angle of 30^o. After walking for

100meters towards the foot of the building he stopped at point B where he observed it again

at an angle of 60^o. Find the height of the building

4.  Find the value of , given that ½ sin = 0.35 for 0^o ≤ θ ≤ 360^o  

5.  A man walks from point A towards the foot of a tall building 240 m away. After covering

180m, he observes that the angle of elevation of the top of the building is 45^o. Determine

the angle of elevation of the top of the building from A

6.  The table below gives a field book showing the results of a survey of a section of a piece of land

between A and E. All measurements are in metres.

 (a) Draw a sketch of the land.

 (b) Calculate the area of this piece of land.  

7.  Solve for x in 2 Cos2x⁰ = 0.6000 0⁰≤ x ≤ 360⁰.

8.  Wangechi whose eye level is 182cm tall observed the angle of elevation to the top of

  her house to be 32º from her eye level at point A. she walks 20m towards the house

on a straight line to a point B at which point she observes the angle of elevation to the

top of the building to the 40º. Calculate, correct to 2 decimal places the

 a)distance of A from the house

 b) The height of the house

9.  Given that cos A = ⁵/₁₃ and angle A is acute, find the value of:-

2 tan A + 3 sin A  

10.  Given that tan 5° = 3 + 5, without using tables or a calculator, determine tan 25°, leaving

your answer in the form a + b c

11.  A student whose eye level is 182cm from the ground observed the top of their house at

an angle of elevation of 32^o at point A. She walked for 20m towards the house along a

straight road to a point B, where she observed the top of the building again at an angle of

elevation of 40^o. Calculate correct to 2 decimal places the:-

  (a) Distance of A from the house

  (b) The height of the house

12.  Given that tan x = 5, find the value of the following without using mathematical tables or

calculator: 12

 (a) Cos x

 (b) Sin²(90-x)

13.  If tan θ =⁸/₁₅, find the value of Sinθ – Cosθ without using a calculator or table

  Cosθ + Sinθ

The trigometric ratio 1 Answers

1.  

Tan 30^o = x x

  100+ y

x = (100 + y) tan 30^o

(100 + y) tan 30^{o =} y tan 60^o

Tan 60^o = x = x = y tan 60^o

  ^y

 (100 + y) 0.5774 = 1.1732y

57.74 = 1.155y

y = 57.74

  1.155

y = 49.99  50m

x = 50 tan 60

 x = 86.6m

2.  Sin = 0.70

  = 44.43°, 135.57°  

3.  (a) (i) Area of triangle A¹B¹C¹ = ½ X 4 X 4

= 8 sq. units

 (b) (ii) Reflection in the line y = x

(c) combine transformation = 0 1 2 0

1 0 0 2

   0 2

 2 0

 Det 0 2 0 –2 x 2 = – 4

2 0

Inverse transformation = – ¼ 0 2 = 0 ^-1/₂

2 0 ^-1/₂ 0

4.  

Tan 45 = AB

60

AB = 45

Tan θ = 45

240

= 0. 1875

θ = 10.62^o

5.  

 Area A: ½ x 25 (33 + 21) = 675

  Area B: ½ x 40 (21 x 42) = 1260

  Area C: ½ x 30 x 42 = 630

 Area D: ½ x 25 x 40 = 500

  Area E: ½ x 5 (40 + 25) = 162.5

  Area F: ½ X 60 (25 + 36) = 1830

 Area G: ½ x 5 x 36 = 90 √

= 5,147.5m²

6.  ∴ Philip takes 10 days.

2Cos 2x = 0.600

Cos 2x = 0.3000

2x = 72.5^o, 287.5

 x = 36.25⁰, 143.75

7.  a)

h

A

Tan32 = h

20 + x

h = (20 +x) tan32º = 12.498 + 0.6249x

tan 40º = ^h/_x

h = x tan 40º = 0.8391x

0.8391x = 12.498 + 0.6249x

0.8391x – 0.6249x = 12.498

  0.2142x = 12.498

x = 12.498 = 58.35m

  0.2142

The distance of A from the house

= (20 + 58.35)m = 78.35

b) h = x tan 40º = 58.35 x 0.8391 = 48.96m

 The total height of the house

= 1.82m + 48.96m = 50.78m

11.   tan 32^oc = h

  20 + x

h = (20 + x) tan 32^o

tan 40^o = h

  x

h = tan 40^o

 x tan 40^o = (20 + x) tan 32^o

0.8391x = (20 + x) 0.6249

0.8391x = 12.498 + 0.6249x

0.8391x – 0.6249x = 12.498

 x = 58.35m

20 + 58.35 = 78.35m

(b) The height of the house

Tan 40^o = h = h = 58.35 tan 40^o

58.35

h = 58.35 x 0.8391

h = 48.96 + 1.82

h = 50.78

12.  24 = 2R ⇒R = 16.15 cm

Sin 48

 Area = 3.14 x 16.15²

= 819.26 cm²

 Hyp = 5² + 12²

 = 13

cos x = ¹²/₁₃

(b) Sin2990-x)

= (¹²/₁₃)²= ¹⁴⁴/₁₆₉

]14.  Tan  = ⁸/_{15 C}

 17

B 15 A

 AB² = 8² + 15²

AB = 289 = 17

Sin  = ⁸/₁₇, cos  = ¹⁵/₁₇

Sin  – cos  = ⁸/₁₇ – ¹⁵/₁₇ = ^-7/₁₇ x ¹⁷/₂₃

Cos  + sin 
¹⁵/₁₇ + ⁸/₁₇

= ^-7/₂₃