Example 1 :
Determine whether the following measurements produce one triangle, two triangles or no triangle:
∠B = 88°, a = 23, b = 2Sine formula
Solve if solution exists.
Solution :
Sine formula :
a/sinA = b/sinB = c/sinC
23/sinA = 2/sin88° = c/sinC
23/sinA = 2/sin88°
sinA = (23/2)sin 88°
sinA = 11.5sin 88°
The maximum value of sinA is 1. So, it is not possible.
Hence the given measurements will not produce a triangle.
Example 2 :
If the sides of a triangle ABC are a = 4, b = 6 and c = 8, then show that 4cosB + 3cosC = 2.
Solution :
Cosine formula :
cosA = (b2 + c2 - a2)/2bc ----(1)
cosB = (a2 + c2 - b2)/2ac ----(2)
cosC = (a2 + b2 - c2)/2ab ----(3)
From (2),
cosB = (a2 + c2 - b2)/2ac
= (42 + 82 - 62)/[2(4)(8)]
= (16 + 64 - 36)/64
= 44/64
= 11/16
From (3),
cosC = (a2 + b2 - c2)/2ab
= (42 + 62 - 82)/[2(4)(6)]
= (16 + 36 - 64)/48
= -12/48
= - 1/4
4cosB + 3cosC = 4(11/16) + 3(-1/4)
= (11/4) - (3/4)
= (11 - 3)/4
= 8/4
= 2
Example 3 :
In a triangle ABC, if a = √3 − 1, b = √3 + 1 and C = 60°, find the other side and other two angles
Solution :
cosC = (a2 + b2 - c2)/2ab
Substitute a = √3 - 1 and b = √3 + 1.
a2 = (√3 - 1)2
= 3 + 1 - 2√3
a = 4 - 2√3
b = √3 + 1,
b2 = (√3 + 1)2
= 3 + 1 + 2√3
= 4 + 2√3
Substituting these values in cosine formula,
cos60 = (4 - 2√3 + 4 + 2√3 - c2)/[2(√3 − 1)(√3 + 1)]
1/2 = (8 - c2)/[2(2)]
1/2 = (8 - c2)/4
Multiply each side by 4.
2 = 8 - c2
c2 = 8 - 2
c2 = 6
c = √6
Sine formula :
a/sinA = b/sinB = c/sinC
(√3 - 1)/sinA = (√3 + 1)/sinB = √6/sin60°
(√3 + 1)/sinB = √6/(√3/2)
(√3 + 1)/sinB = 2√6/√3
(√3 + 1)/sinB = 6√2/3
(√3 + 1)/sinB = 2√2
sinB = (√3 + 1)/2√2
sinB = (√3/2)(1/√2) + (1/2)(1/√2)
= sin60°cos45° + cos60°sin45°
= sin(60° + 45°)
= sin105°
∠B = 105°
In triangle ABC,
∠A + ∠B + ∠C = 180°
Substitute.
∠A + 105° + 60° = 180°
∠A + 165° = 180°
∠A = 15°
Therefore,
c = √6, ∠A = 15°, ∠B = 105°
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Nov 02, 24 11:58 PM
Oct 30, 24 10:07 AM
Oct 29, 24 06:24 AM