15.2.3 Trigonometry, PT3 Focus Practice


Question 10:
A technician needs to climb a ladder to repair a street lamp as shown in Diagram below.
(a) What is the length, in m, of the ladder?
(b) Find the height, in m, of the lamp post.

Solution:


(a)

cos 70 o = 1 h h= 1 cos 70 o h= 1 0.342 h=2.924 m Hence, the length of the ladder =2.924 m

(b)
tan 70 o = T 1 T=tan 70 o ×1    =2.747×1    =2.747 m The height of the lamp post =2.747+1.2 =3.947 m


15.2.2 Trigonometry, PT3 Focus Practice


Question 6:
In diagram below, AEC and BCD are straight lines. E is the midpoint of AC.

Given cosx= 5 13  and siny= 3 5
(a) find the value of tan x.
(b) Calculate the length, in cm, of BD.

Solution:

(a) Given cos x= 5 13 , therefore BC=5, AB=13 AC= 13 2 5 2  = 16925  = 144  =12 cm tan x= AC BC = 12 5


(b) For ΔDCE: siny= 3 5 EC DE = 3 5 EC 10 = 3 5 EC= 3 5 ×10=6 cm D C 2 = 10 2 6 2    =64   DC=8 cm For ΔABC: AC=2×6=12 cm tanx= 12 5 12 CB = 12 5 CB=5 cm BD=DC+CB =8 cm + 5 cm =13 cm



Question 7:
In diagram below, T is the midpoint of the line PR.

(a) Find the value of tan xo.
(b) Calculate the length, in cm, of PQ.

Solution:
(a) T R 2 = 13 2 12 2   =169144   =25 TR= 25  =5 cm tan x o = 12 5

(b) PR=2×5 cm  =10 cm P Q 2 = 10 2 8 2    =10064    =36 PQ= 36  =6 cm

Question 8:
In diagram below, ABE and DBC are two right-angled triangles ABC and DEB are straight lines.


It is given that cos y o = 3 5 .
(a) Find the value of tan xo.
(b) Calculate the length, in cm, of DE.

Solution:
(a) tan x o = 7 24

(b) cos y o = BC 20    3 5 = BC 20 BC= 3 5 ×20  =12 cm B D 2 = 20 2 12 2   =400144   =256 BD= 256  =16 cm DE=167   =9 cm

Question 9:
Diagram below shows a vertical pole, PQ. At 2.30 p.m. and 5.00 p.m., the shadow of the pole falls on QR and QS respectively.
Calculate
(a) the height, in m, of the pole.
(b) the value of w.

Solution:
(a)
tan  55 o = Height of the pole 3.2 Height of the pole=tan  55 o ×3.2 =1.428×3.2 =4.57 m

(b)
tan w= 4.57 3.20+2  = 4.57 5.20  =0.879    w= 41 o 18'


15.2.1 Trigonometry, PT3 Focus Practice


15.2.1 Trigonometry, PT3 Focus Practice

Question 1:
Diagram below shows a right-angled triangle ABC.



It is given that  cos x o = 5 13 , calculate the length, in cm, of AB.

Solution:
cos x o = AB AC cos x o = 5 13 AB 39 = 5 13 AB= 5 13 ×39  =15 cm



Question 2:
In the diagram, PQR and QTS are straight lines.


It is given that tany= 3 4 , calculate the length, in cm, of RS.

Solution:
In  PQT, tany= PQ QT 3 4 = 6 QT QT=6× 4 3  =8 cm In QRS, QS=8+8=16 cm R S 2 = 12 2 + 16 2   pythagoras' Theorem     =144+256   =400 RS= 400  =20 cm



Question 3:
In the diagram, PQR is a straight line.

It is given that   cos x o = 3 5 , hence sin yo =

Solution:
cos x o = PQ PS PQ 10 = 3 5 PQ= 3 5 ×10  =6 cm QR=PRPQ =216 =15 cm


Q S 2 = 10 2 6 2  pythagoras' Theorem     =10036    =64 QS= 64  =8 cm R S 2 = 15 2 + 8 2    =225+64    =289 RS= 289  =17 cm sin y o = 15 17


Question 4:
Diagram below consists of two right-angled triangles.

Determine the value of cos xo.

Solution:
AC= 13 2 12 2  = 25  =5 cm CD= 5 2 3 2  = 16  =4 cm cos x o = CD AC   = 4 5


Question 5:
Diagram below consists of two right-angled triangles ABE and DBC.
ABC and EBD are straight lines.



It is given that sin x o = 5 13  and cos y o = 3 5 .
(a) Find the value of tan xo.
(b)   Calculate the length, in cm, of ABC.

Solution:
(a)
sin x o = 5 13 , DC=13 cm BC= 13 2 5 2  = 144  =12 cm Thus, tan x o = 5 12

(b)
cos y o = AB 15    3 5 = AB 15 AB=9 cm Thus, ABC=9+12  =21 cm

15.1 Trigonometry


15.1 Trigonometry
 
15.1.1 Trigonometrical Ratios of an Acute Angle

1.    


2.   Hypotenuse is the longest side of the right-angled triangle which is opposite the right angle.
3.   Adjacent side is the side, other than the hypotenuse, which has direct contact with the given angle, θ.
4.   Opposite side is the side which is opposite the given angle, θ.
5.   In a right-angled triangle,

     sinθ= opposite side hypotenuse      cosθ= adjacent side hypotenuse   tanθ= opposite side adjacent side      

6.
When the size of an angle θ increases from 0o to 90o,

· sin θ increases.
· cos θ increases.
· tan θ increases.

7.   The values of sin θ, cos θ and tan θ remain the same even though the size of the triangle has changed.
Example:

Find the sine, cosine and tangent of the give angle, θ, in triangle ABC.

Solution:
sin θ = B C A B = 8 17 cos θ = A C A B = 15 17 t a n θ = B C A C = 8 15
 

15.1.2  Values of Tangent, Sine and Cosine
1.   The values of the trigonometric ratios of 30o, 45o and 60o (special angle) are as below.
 




2.   1 degree is equal to 60 minutes.
1o= 60’

3.   A scientific calculator can be used to find the value of the sine, cosine or tangent of an angle.
Example:
sin 40.6o = 0.5954
 
Calculator Computation
    Press [sin] [40.6] [=] 0.595383839
4.   Given the values of sine, cosine and tangent, we can find the angles using a scientific calculator.
Example:
tan x = 1.7862
  x = 67o30’
 
Calculator Computation
Press [shift] [tan][1.7862]
 [=][o’’’]  67o30’