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The other cases can be proved in an identical manner.

Find A ^ :

  1. a 2 = b 2 + c 2 - 2 b d cos A ^ cos A ^ = b 2 + c 2 - a 2 2 b c = 8 2 + 5 2 - 7 2 2 · 8 · 5 = 0 , 5 A ^ = arccos 0 , 5 = 60

The cosine rule

  1. Solve the following triangles i.e.  find all unknown sides and angles
    1. ABC in which A ^ = 70 ; b = 4 and c = 9
    2. XYZ in which Y ^ = 112 ; x = 2 and y = 3
    3. RST in which RS = 2 ; ST = 3 and RT = 5
    4. KLM in which KL = 5 ; LM = 10 and KM = 7
    5. JHK in which H ^ = 130 ; JH = 13 and HK = 8
    6. DEF in which d = 4 ; e = 5 and f = 7
  2. Find the length of the third side of the XYZ where:
    1. X ^ = 71 , 4 ; y = 3 , 42  km and z = 4 , 03  km
    2. ; x = 103 , 2  cm; Y ^ = 20 , 8 and z = 44 , 59  cm
  3. Determine the largest angle in:
    1. JHK in which JH = 6 ; HK = 4 and JK = 3
    2. PQR where p = 50 ; q = 70 and r = 60

The area rule

The Area Rule

The area rule applies to any triangle and states that the area of a triangle is given by half the product of any two sides with the sine of the angle between them.

That means that in the D E F , the area is given by: A = 1 2 D E · E F sin E ^ = 1 2 E F · F D sin F ^ = 1 2 F D · D E sin D ^

In order show that this is true for all triangles, consider A B C .

The area of any triangle is half the product of the base and the perpendicular height. For A B C , this is: A = 1 2 c · h . However, h can be written in terms of A ^ as: h = b sin A ^ So, the area of A B C is: A = 1 2 c · b sin A ^ .

Using an identical method, the area rule can be shown for the other two angles.

Find the area of ABC:

  1. ABC is isosceles, therefore AB = AC = 7 and C ^ = B ^ = 50 . Hence A ^ = 180 - 50 - 50 = 80 . Now we can use the area rule to find the area:

    A = 1 2 c b sin A ^ = 1 2 · 7 · 7 · sin 80 = 24 , 13

The area rule

Draw sketches of the figures you use in this exercise.

  1. Find the area of PQR in which:
    1. P ^ = 40 ; q = 9 and r = 25
    2. Q ^ = 30 ; r = 10 and p = 7
    3. R ^ = 110 ; p = 8 and q = 9
  2. Find the area of:
    1. XYZ with XY = 6  cm; XZ = 7  cm and Z ^ = 28
    2. PQR with PR = 52  cm; PQ = 29  cm and P ^ = 58 , 9
    3. EFG with FG = 2 , 5  cm; EG = 7 , 9  cm and G ^ = 125
  3. Determine the area of a parallelogram in which two adjacent sides are 10 cm and 13 cm and the angle between them is 55 .
  4. If the area of ABC is 5000 m 2 with a = 150  m and b = 70  m, what are the two possible sizes of C ^ ?

Summary of the trigonometric rules and identities

Pythagorean Identity Ratio Identity
cos 2 θ + sin 2 θ = 1 tan θ = sin θ cos θ
Odd/Even Identities Periodicity Identities Cofunction Identities
sin ( - θ ) = - sin θ sin ( θ ± 360 ) = sin θ sin ( 90 - θ ) = cos θ
cos ( - θ ) = cos θ cos ( θ ± 360 ) = cos θ cos ( 90 - θ ) = sin θ
Sine Rule Area Rule Cosine Rule
Area = 1 2 bc cos A a 2 = b 2 + c 2 - 2 b c cos A
sin A a = sin B b = sin C c Area = 1 2 ac cos B b 2 = a 2 + c 2 - 2 a c cos B
Area = 1 2 ab cos C c 2 = a 2 + b 2 - 2 a b cos C

Exercises

  1. Q is a ship at a point 10 km due South of another ship P. R is a lighthouse on the coast such that P ^ = Q ^ = 50 . Determine:
    1. the distance QR
    2. the shortest distance from the lighthouse to the line joining the two ships (PQ).
  2. WXYZ is a trapezium (WX XZ) with WX = 3  m; YZ = 1 , 5  m; Z ^ = 120 and W ^ = 30
    1. Determine the distances XZ and XY.
    2. Find the angle C ^ .
  3. On a flight from Johannesburg to Cape Town, the pilot discovers that he has been flying 3 off course. At this point the plane is 500 km from Johannesburg. The direct distance between Cape Town and Johannesburg airports is 1 552 km. Determine, to the nearest km:
    1. The distance the plane has to travel to get to Cape Town and hence the extra distance that the plane has had to travel due to the pilot's error.
    2. The correction, to one hundredth of a degree, to the plane's heading (or direction).
  4. ABCD is a trapezium (ie. AB CD). AB = x ; B A ^ D = a ; B C ^ D = b and B D ^ C = c . Find an expression for the length of CD in terms of x , a , b and c .
  5. A surveyor is trying to determine the distance between points X and Z. However the distance cannot be determined directly as a ridge lies between the two points. From a point Y which is equidistant from X and Z, he measures the angle X Y ^ Z .
    1. If XY = x and X Y ^ Z = θ , show that XZ = x 2 ( 1 - cos θ ) .
    2. Calculate XZ (to the nearest kilometre) if x = 240  km and θ = 132 .
  6. Find the area of WXYZ (to two decimal places):
  7. Find the area of the shaded triangle in terms of x , α , β , θ and φ :

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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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