15.2 Trigonometry, PT3 Focus Practice


15.2 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=PR−PQ =21−6 =15 cm
Q S 2 = 10 2 − 6 2 ← pythagoras' Theorem        =100−36        =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      =5cm CD= 5 2 − 3 2      = 16      =4cm 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=13cm BC= 13 2 − 5 2      = 144      =12cm 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


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