To solve this problem, we first need to establish a relationship between the work rates of the women, boys, and the man.

Let the amount of work done by 1 woman in one day be W, and the amount of work done by 1 boy in one day be B.

From the information provided:

4 women = 6 boys in terms of work rate. Therefore,

4W = 6B → W = (6/4)B → W = (3/2)B.

Now let’s determine the combined work rates:

– The work done by 4 women in one day is 4W = 4 * (3/2)B = 6B.

– The work done by 6 boys in one day is 6B.

This means 4 women can complete the same amount of work in the same time as 6 boys.

Next, let’s find the amount of work done by a man:

A man can finish the work in 60 days, so if we denote the total work as 1 unit, then the work rate of the man (M) is:

M = 1/60 units of work per day.

Now, we need to relate the man’s work rate to the women’s. Since we established that 4 women can do the work equivalent to 6 boys, we need to solve for how many days 5 women will take to finish the work.

From our earlier calculation:

1 man’s work rate =

To find the length of AB in triangle ABC, which is congruent to triangle PQR, we can use the properties of right triangles.

Given:

– Triangle ABC is a right triangle with angle A = 90°.

– BC = 13 cm (hypotenuse).

– Triangle PQR is also a right triangle with angle P = 90° and PR = 5 cm (hypotenuse).

Since triangles ABC and PQR are congruent, it means all corresponding sides and angles are equal.

Thus, if PR = 5 cm, BC in triangle ABC must also correspond to the hypotenuse of triangle PQR. Given that BC = 13 cm, we can use the Pythagorean theorem to solve for AB.

In triangle ABC:

[ AB^2 + AC^2 = BC^2 ]

Let ( AC = x ). Since PR is also given to be equal to the corresponding sides of triangle ABC (which means corresponding sides in congruent triangles are equal):

– Let ( PQ = AB ) and ( QR = AC ).

– Therefore, ( AB ) corresponds to ( PR = 5 ) cm.

Now we have:

1. ( BC^2 = AB^2 + AC^2 )

[ 13^2 = AB^2 + AC^2 ]

[ 169 = AB^2 + AC^2 ]

2. Using the fact that ( AC ) corresponds to

To simplify the expression ( cos(36^circ + A) cos(36^circ – A) + cos(54^circ + A) cos(54^circ – A) ), we can use the cosine addition formula:

[

cos(x + y) cos(x – y) = frac{1}{2} left( cos(2x) + cos(2y) right)

]

Applying this formula to each term:

1. For ( cos(36^circ + A) cos(36^circ – A) ):

[

cos(36^circ + A) cos(36^circ – A) = frac{1}{2} left( cos(72^circ) + cos(2A) right)

]

2. For ( cos(54^circ + A) cos(54^circ – A) ):

[

cos(54^circ + A) cos(54^circ – A) = frac{1}{2} left( cos(108^circ) + cos(2A) right)

]

Now, we can combine these results:

[

cos(36^circ + A) cos(36^circ – A) + cos(54^circ + A) cos(54^circ

To solve this problem, we can set up a system of equations based on the information provided.

Let:

– ( w ) = earnings of 1 woman per day

– ( g ) = earnings of 1 girl per day

From the first scenario:

5 women and 9 girls earn ₹18,720 in 9 days.

The equation can be set up as:

[

9(5w + 9g) = 18720

]

This simplifies to:

[

5w + 9g = frac{18720}{9} = 2080 quad text{(Equation 1)}

]

From the second scenario:

9 women and 16 girls earn ₹52,080 in 14 days.

Setting up the equation:

[

14(9w + 16g) = 52080

]

This simplifies to:

[

9w + 16g = frac{52080}{14} = 3720 quad text{(Equation 2)}

]

Now we have the system of equations:
1. ( 5w + 9g = 2080 )
2. ( 9w + 16g = 3720 )

We can solve these equations simultaneously.

From Equation 1, we can express ( g ) in terms of ( w ):

[

9g = 2080 – 5w

]

[

g =