9.3 Trigonometry II, SPM Paper 1 (Short Questions)


Question 4:

In the diagram above, WZY  is a straight line.  X Y Z = 90 o , X W Z = 30 o and WZ = XZ = 30 cm. Find the length of XY.

Solution:
WXZ=XWZ= 30 o XZY= 30 o + 30 o = 60 o sinXZY= XY XZ sin 60 o = XY 30 XY=sin 60 o ×30 XY=25.98cm


Question 5:

In the diagram above, PQS is a right angle triangle. Given that SR = 6cm, PQ = 12 cm and 5SR = 2PS. Find the value of cos α and tan β.

Solution:
5SR=2PS PS= 5 2 SR PS= 5 2 ( 6 ) PS=15 cm cosα= PQ PS cosα= 12 15 = 4 5 In  PQS, using Pythagoras’ Theorem, QS= P S 2 P Q 2 QS= 15 2 12 2 =9 cm tanβ=tanPSQ Since  90 <β< 180 (in quadrant II), tanβ is negative tanβ= PQ QS tanβ= 12 9 = 4 3


Question 6:

In the diagram above, ADC is a straight line, if  sin q = 3 5 and tan p = 1 2 . Find the distance of AC.

Solution:
Given sin q = B D A B = 3 5 B D 30 = 3 5 B D = 3 5 × 30 B D = 18 c m In A B D , using Pythagoras’ Theorem, A D = A B 2 B D 2 A D = 30 2 18 2 = 24 c m Given tan p = B D D C = 1 2 18 D C = 1 2 D C = 36 c m Hence, distance of A C = 24 + 36 = 60 c m .

4 Comments to “9.3 Trigonometry II, SPM Paper 1 (Short Questions)”

  1. Hi, for question 5 the answer for cos alpha is 4/5 not 3/5. Please correct the simplification.

    1. Dear Gurdit Singh,
      thanks for pointing out our mistake, correction had been made accordingly.

  2. question 5 cos alpha should be 4/5 and not 3/5 as 12/15(which is correct) equals to 4/5

    1. Dear donald,
      thanks for pointing out our mistake, correction had been made accordingly.

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