Let OP be the tower of height h (say) and A and B be the two positions on the horizontal line through O, such that
?OAP = ?, ?OBP = ? and OB = x
In ?OBP, Use the trigonometry formula
Tan? = P/B = Perpendicular distance / Base distance
Tan? = OP/OB
? OB = OP/Tan?
? OB = OP Cot?
Put the value of OB and OP , We will get
x = h Cot ?...............(1)
In ?OAP, Similarly
Tan? = OP/OA
? OA = OP/ Tan?
? OA = OP Cot ?
Put the value of OA and OP
? a + x = h Cot ?
? x = h Cot ? - a ............(2)
From equation (1) and (2)
? h Cot ? = h Cot ? - a
? a = h Cot ? - h Cot ?
? a = h (Cot ? - Cot ?)
? a = h (Cos ?/ Sin ? - Cos ? / Sin ? )
? a = h( (Cos ? Sin ? - Cos ? Sin ? ) /Sin ? Sin ? )
? a = h( Sin(? - ?) / Sin ? Sin ?)
? h = a Sin ? Sin ?/ Sin(? - ?)
There are 3A's 2N's and one B. We have to find the total number of arrangements of 6 letters out of which 3 are alike of one kind, 2 are alike of second kind, thus the total number of words
= 6! / (3! 2!) = 60
S.P. | = P.W. of Rs. 2200 due 1 year hence | |||||
|
||||||
= Rs. 2000. |
∴ Gain = Rs. (2000 - 1950) = Rs. 50.
Since 39 persons can repair a road by working 5 hours in a day in 12 days.
? 1 persons can repair a road by working 5 hours in a day in 39 x 12 days.
? 1 persons can repair a road by working 1 hours in a day in 39 x 12 x 5 days.
? 30 persons can repair a road by working 1 hours in a day in 39 x 12 x 5 / 30 days.
? 30 persons can repair a road by working 6 hours in a day in 39 x 12 x 5 / 30 x 6 days.
? 30 persons can repair a road by working 6 hours in a day in 39 x 2 x 1 / 6 days.
? 30 persons can repair a road by working 6 hours in a day in 39 x 1 x 1 / 3 days.
? 30 persons can repair a road by working 6 hours in a day in 13 days.
A : C = 100 : 72.
∴ | B | = | ❨ | B | x | A | ❩ | = | ❨ | 80 | x | 100 | ❩ | = | 10 | = | 100 | = 100 : 90. |
C | A | C | 100 | 72 | 9 | 90 |
∴ B can give C 10 points.
Then, | 30 | - | 30 | = 3 |
x | 2x |
⟹ 6x = 30
⟹ x = 5 km/hr.
Let the required weight be w kg.
Less length, Less weight (Direct proportion)
45 : 12 :: 171 : w
? w = (12 x 171)/45 = 45.6 kg
Let the sum be Rs 'y' , so amount = 2y
Simple interest =Rs y
Let R be the rate of interest,
R= (100 x SI)/(P x T) = (100 x y) / (y x 8) = 12.5 %
where SI= Simple Interest
P = Principal
T = Time
now, the needed amount = Rs 4y
since SI = Rs (4x-x)= Rs 3y
since T= (100 x SI)/(P x R)
= (100 x 3y )/(y x 125)= 24 yr
Let us assume the first Odd number is a.
then 2nd odd number = a + 2
and 3rd odd number = a + 4
According to question,
Sum of three consecutive odd numbers is 20 more than the first of these numbers;
a + (a + 2) + (a + 4) = a + 20
? a + a + 2 + a + 4 - a = 20
? 2a + 6 = 20
? 2a + 6 = 20
? 2a = 20 - 6
? 2a = 14
? a = 7
The First odd number a = 7
The second odd number = a + 2 = 7 + 2 = 9
Second odd number is middle number = 9
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.