LAW OF SINES APPLICATION PROBLEMS EXAMPLES

About "Law of Sines Application Problems Examples"

Law of Sines Application Problems Examples :

Here we are going to see some example problems based on law of sines.

Law of Sines Application Problems Examples - Examples

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 + c− b2)/2ac

(a2 + c− 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 + b− c2)/2ab

(a2 + b− 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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