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
(a + b)/(a − b) = tan (A + B)/2 cot (A − B)/2
(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
Question 2 :
In a triangle ABC, prove that (a2 − b2 + c2) tanB = (a2 + b2 − c2) tanC.
(a2 − b2 + c2) tanB = (a2 + b2 − c2) tanC.
(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)
(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)
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.
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
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
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
= (12/√3) ⋅ (√3/√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.
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
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
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