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

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