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Ask QuestionPosted by Taniya 5 years, 1 month ago
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Avatar ? 5 years, 1 month ago
Posted by Shreedhar J 5 years, 1 month ago
- 2 answers
Yogita Ingle 5 years, 1 month ago
Let a be any positive integer and b = 4. Then by Euclid’s algorithm,
a = 4q + r for some integer q ≥ 0, and r = 0, 1, 2, 3
So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3 because 0 ≤ r < 4
Now, 4q i.e., 2(2q) is an even number
∴4q + 1 is an odd number.
4q + 2 i.e., 2(2q + 1) which is also an even number.
∴ (4q + 2) + 1 = 4q + 3 is an odd number.
Thus, we can say that any odd integer can be written in the form 4q + 1 or 4q + 3 where q is some integer.
Shreedhar J 5 years, 1 month ago
Posted by A Mariyappan 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
1445 = 1190*1 + 255
1190 = 255*4 + 170
255 = 170*1 + 85
170 = 85*2 + 0
So, now the remainder is 0, then HCF is 85
Now,
85 = 255 - 170
(1445 - 1190) - (1190 - 255*4)
⇒ 1445 - 1190 - 1190 + 255*4
⇒ 1445 - 1190*2 + (1445 - 1190)*4
⇒ 1445 - 1190*2 + 1445*4 - 1190*4
⇒ 1445*5 - 1190*6
⇒ 1190*(- 6) + 1445*5
1190m + 1445n , where m = - 6 and n = 5
Posted by Lucky Sharma 5 years, 1 month ago
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Posted by Geeta Jadav 5 years, 1 month ago
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Posted by P L 5 years, 1 month ago
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Posted by Arpita Pathak 5 years, 1 month ago
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Posted by Rishabh Shukla 5 years, 1 month ago
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Posted by Archana Kumari 5 years, 1 month ago
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Saloni Pathak 5 years, 1 month ago
Posted by Princess Of God 5 years, 1 month ago
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Kashish . 5 years, 1 month ago
Posted by Siddhi Pateriya 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:
(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
Solution:
Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively.

(i) From the given condition,
ΔABC ~ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
Since, ∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ΔACD and ΔFGH,
∠A = ∠F
∠ACD = ∠FGH
∴ ΔACD ~ ΔFGH (AA similarity criterion)
⇒CD/GH = AC/FG
(ii) In ΔDCB and ΔHGE,
∠DCB = ∠HGE (Already proved)
∠B = ∠E (Already proved)
∴ ΔDCB ~ ΔHGE (AA similarity criterion)
(iii) In ΔDCA and ΔHGF,
∠ACD = ∠FGH (Already proved)
∠A = ∠F (Already proved)
∴ ΔDCA ~ ΔHGF (AA similarity criterion)
Posted by Jahnavi Velagapudi 5 years, 1 month ago
- 2 answers
Yangzee Sherpa 5 years, 1 month ago
Gaurav Seth 5 years, 1 month ago
Let us consider that 3root2 is a rational number. It can be written in the form p/q (p and q are co primes)
p/q = 3root2
p/3q = root2
Now,
p/3q = integer/interger
= rational number
But, this contradicts the fact that root2 is irrational.
Therefore, our assumption that 3root2 is rational is WRONG.
Hence, 3root2 is an irrational number.
Posted by Rajesh Singh 5 years, 1 month ago
- 3 answers
Posted by Mrunal Keote 5 years, 1 month ago
- 1 answers
Posted by Prãvëëñ Yâdåv 5 years, 1 month ago
- 1 answers
Gaurav Seth 5 years, 1 month ago
If Tan A = 3/4, then find the other trigonometric ratio of angle A.
sol) Tan A = Opposite side = 3
Adjacent side 4
.^., opposite side : adjacent side = 3:4
For angle A,
opposite side = BC=3k
Adjacent side =AB=4k(where k is any +ve no)
Now, we have in triangle ABC( by Pythagoras theorem)

AC^2 = AB^2 + BC^2
=>(3k)^2 + (4k)^2 = 25k^2
AC = √25k^2
=> 5k = Hypotenuse
Now,we can easily write the other ratios of trigonometry
sin A = 3k = 3
5k 5
cos A = 4k = 4
5k 5
cosec A = 1 = 5
sin A 3
sec A = 1 = 5
cos A 4
cot A = 1 = 4
tan A 3
Posted by Divyanshu Yadav 5 years, 1 month ago
- 1 answers
Gaurav Seth 5 years, 1 month ago
Let the speed of car at A be x kmph
and the speed of car at B be y kmph
As per the question,
5x - 5y = 100
x - y = 20 ... (1)
and x + y = 100 ... (2)
Solving (1) and (2), we get,
x = 60 and y = 40
Speed of the car at A = 60 kmph
Speed of the car at B = 40 kmph
Posted by Ritika Ritika 5 years, 1 month ago
- 3 answers
Prabhat Mishra 5 years, 1 month ago
Posted by Pragya Bharti 5 years, 1 month ago
- 1 answers
Gaurav Seth 5 years, 1 month ago
2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1
=2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1
=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2
-2sin2θcos2θ]+1
The algebraic identity
a3 + b3 = (a+b)3 - 3ab(a+b) and
a2 + b2 = (a+b)2 - 2ab
are used in the above step where
a = sin2θ and b = cos2θ.
writing sin2θ + cos2θ = 1, we have
= 2[1-3 sin2θ cos2θ] -3[-2 sin2θcos2θ] + 1
= 2-6 sin2θcos2θ -3 + 6 sin2θcos2θ + 1
= -3+3=0
Posted by Sudha Sudha 5 years, 1 month ago
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Priyanshu Kumar 5 years, 1 month ago
Posted by Neha Ramesh 5 years, 1 month ago
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Himanshu Nagaich 5 years, 1 month ago
Posted by Khusbhoo Kumari 5 years, 1 month ago
- 2 answers
Gaurav Seth 5 years, 1 month ago
A n s w e r :
=> (sin30+cos30) - (sin60+cos60)
sin30 = 1/2
sin60= √3/2
cos30= √3/2
cos60= 1/2
=> (1/2+√3/2) - (√3/2+1/2)
=> 1+√3/2 - √3+1/2
=> 1+1+√3-√3 /2
=> 2/2
=>1
Posted by Chetan Soni 5 years, 1 month ago
- 2 answers
Harini Reddy 5 years, 1 month ago
Posted by Vashu Tyagi 5 years, 1 month ago
- 1 answers
Geeta Jadav 5 years, 1 month ago
Posted by Durga Saidev Harsha Kasireddy 5 years, 1 month ago
- 5 answers

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Shubham Solanki 5 years, 1 month ago
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