The cosine law can be used which is not a right triangle.
a^{2} = b^{2} + c^{2} - 2bc cos A
b^{2} = c^{2} + a^{2} - 2ac cos B
c^{2} = a^{2} + b^{2} - 2ab cos C
Problem 1 :
Calculate the perimeter of triangle ABC.
Solution:
By using cosine rule,
a^{2} = b^{2} + c^{2} - 2bc cos(A)
a^{2} = 8^{2} + 9^{2} - 2(8)(9) cos(30°)
a^{2} = 64 + 81 - 72(√3)
a^{2} = 145 - 124.7
a^{2} = 20.29
a = √20.29
a = 4.5 cm
Perimeter = a + b + c
= 4.5 + 8 + 9
= 21.5 cm
So, the perimeter of triangle ABC is 21.5 cm.
Problem 2 :
Boat A is 16 km from a lighthouse on a bearing of 055°. Boat B is 11 km from the same lighthouse on a bearing of 152°. Calculate the distance between the two boats.
Solution:
∠BLA = 152° - 55°
= 87°
By using cosine law,
AB^{2} = LA^{2} + LB^{2} - 2(LA)(LB) Cos ∠BLA
= 16^{2} + 11^{2} - 2(16)(11) Cos (87°)
= 256 + 121 - 352(0.052)
= 377 - 18.42
AB^{2} = 358.58
AB = √358.58
= 18.94 km
So, the distance between the two boat is 18.94 km.
Problem 3 :
Shown is sector OAB.
O is the centre of the circle with radius 9 cm.
A and B are points on the circle.
The length of the chord AB is 10.35 cm.
Find the area of sector OAB.
Solution:
Given, OA = 9 cm, OB = 9 cm and AB = 10.35 cm
By using cosine law,
Problem 4 :
A hot air balloon is flying above two point, standing on the ground at points A and B, 600m apart.
The hot air balloon is 300 m from A and 500 m from B.
(a) Work out the angle of elevation from point B.
(b) How high is the hot air balloon from the ground?
Solution:
a)
500^{2} - x^{2} = 300^{2} - (600 - x)^{2}
250000 - x^{2} = 90000 - x^{2} + 1200x - 360000
1200x = 520000
b.
Problem 5 :
A boat, located at position X is running out of fuel.
There are two ports located at Y and Z.
The boat must refuel as soon as possible.
How much closer is the boat to the port at Y than the port at Z?
Solution:
angle x is equal to 180° - 60° - 45° = 75°
Using the law of sine,
y - z = 25.12 - 20.5
= 4.62 miles
The distance is 4.62 miles.
From X to Z is 4.62 miles closer then X to Y.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM