**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 (a^{2} − b^{2} + c^{2}) tanB = (a^{2} + b^{2} − c^{2}) tanC.

**Solution :**

(a^{2} − b^{2} + c^{2}) tanB = (a^{2} + b^{2} − c^{2}) tanC.

L.H.S

(a^{2} − b^{2} + c^{2}) tanB -----(1)

Cosine formula for cos B = (a^{2} + c^{2 }− b^{2})/2ac

(a^{2} + c^{2 }− b^{2}) = 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

(a^{2} + b^{2} − c^{2}) tanC ---(2)

Cosine formula for cos C = (a^{2} + b^{2 }− c^{2})/2ab

(a^{2} + b^{2 }− c^{2}) = 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) ≤ (s^{3}/27)

Let s-a = x, s-b = y and s-c = z

xyz ≤ (s^{3}/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 s^{2}/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 = 6^{2}/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 = (b^{2} + c^{2} - a^{2})/2bc ---(1)

cos B = (c^{2} + a^{2} - b^{2})/2ac ------(2)

cos C = (a^{2} + b^{2} - c^{2})/2ab

From (1)

b cos A = (b^{2} + c^{2} - a^{2})/2c

a cos B = (c^{2} + a^{2} - b^{2})/2c

b cos A + a cos B = [(b^{2} + c^{2} - a^{2}) + (c^{2} + a^{2} - b^{2})]/2c

b cos A + a cos B = 2c^{2}/2c

b cos A + a cos B = c

Hence proved.

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