Law of Sines Application Problems Examples :
Here we are going to see some example problems based on law of sines.
Question 1 :
In a triangle ABC, prove that
(v) (a + b)/(a − b) = tan (A + B)/2 cot (A − B)/2
Solution :
(a + b)/(a − b) = tan (A + B)/2 cot (A − B)/2
L.H.S
(a + b)/(a − b)
= (2R sin A + 2R sin B)/(2R sin A - 2R sin B)
= (sin A + sin B)/(sin A - sin B)
= [2 sin (A+B)/2 cos (A-B)/2]/[2 cos (A+B)/2 sin (A-B)/2]
= [sin (A+B)/2 cos (A-B)/2]/[cos (A+B)/2 sin (A-B)/2]
= tan (A+B)/2 cot (A-B)/2
R.H.S
Hence proved.
Question 2 :
In a triangle ABC, prove that (a2 − b2 + c2) tanB = (a2 + b2 − c2) tanC.
Solution :
(a2 − b2 + c2) tanB = (a2 + b2 − c2) tanC.
L.H.S
(a2 − b2 + c2) tanB -----(1)
Cosine formula for cos B = (a2 + c2 − b2)/2ac
(a2 + c2 − b2) = 2ac cos B
= 2ac cos B tan B
= 2ac cos B (sin B/cos B)
= 2ac sin B
= 2ac (b/2R)
= abc/R -----(L.H.S)
R.H.S
(a2 + b2 − c2) tanC ---(2)
Cosine formula for cos C = (a2 + b2 − c2)/2ab
(a2 + b2 − c2) = 2ab cos C
= 2ab cos C tan C
= 2ab cos C (sin C/cos C)
= 2ab sin C
= 2ab (c/2R)
= abc/R -----(R.H.S)
Hence proved.
Question 3 :
An Engineer has to develop a triangular shaped park with a perimeter 120 m in a village. The park to be developed must be of maximum area. Find out the dimensions of the park.
Solution :
From heron's formula
Area of triangle = √s (s-a) (s-b) (s-c)
Here s = (a + b + c)/2
The triangle will have maximum area, if the values of (s-a), (s-b) and (s-c) were maximum.
There is a relationship between arithmetic mean and geometric mean.
Geometric mean ≤ Arithmetic mean ---(1)
Let s-a, s-b and s-c are the terms of arithmetic progression and geometric progression.
Arithmetic mean of s-a, s-b and s-c is
= (s-a + s-b + s-c)/3
= (3s - a - b - c)/3
= (3s - (a+b+c))/3
= (3s - 2s) /3
= s/3 -------(1)
Geometric mean of s-a, s-b and s-c is
= [(s-a)(s-b)(s-c)]1/3
By applying those values in (1), we get
[(s-a)(s-b)(s-c)]1/3 ≤ (s/3)
(s-a)(s-b)(s-c) ≤ (s/3)3
(s-a)(s-b)(s-c) ≤ (s3/27)
Let s-a = x, s-b = y and s-c = z
xyz ≤ (s3/27)
If xyz ≤ k, then x = y = z
If s - a = s - b = s - c, then a = b = c.
a + b + c = 120
3a = 120
a = 40
Hence a = 40, b = 40 and c = 40.
Question 4 :
A rope of length 12 m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed
Solution :
If the perimeter of triangle is constant, then the maximum area of triangle will be s2/3√3 and a = b = c.
Given that perimeter of triangle = 12
a + b + c = 12
3a = 12
a = 4
s = (a + b + c)/2 ==> 12/2 ==> 6
Area of triangle = 62/3√3
= 36/3√3
= (12/√3) ⋅ (√3/√3)
= 4√3
Hence the dimensions of the triangle are 4 m ,4 m and 4 m and area of triangle is 4√3 m.
Question 5 :
Derive Projection formula from (i) Law of sines, (ii) Law of cosines.
Solution :
Projection formula :
a = b cos C + c cos B ----(1)
b = c cos A + a cos C
c = a cos B + b cos A
(1) ==> a = 2R sin A
A + B + C = 180
= 2R sin (180 - (B + C))
= 2R sin (B + C)
= 2R (sin B cos C + cos B sin C)
= 2R sin B cos C + 2R sin C cos B
= b cos C + c cos B
So, a= b cos C + c cos B.
Similarly we may prove the other statements.
(ii) Cosine formula
cos A = (b2 + c2 - a2)/2bc ---(1)
cos B = (c2 + a2 - b2)/2ac ------(2)
cos C = (a2 + b2 - c2)/2ab
From (1)
b cos A = (b2 + c2 - a2)/2c
a cos B = (c2 + a2 - b2)/2c
b cos A + a cos B = [(b2 + c2 - a2) + (c2 + a2 - b2)]/2c
b cos A + a cos B = 2c2/2c
b cos A + a cos B = c
Hence proved.
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