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Ask QuestionPosted by Lata Thapa 6 years, 10 months ago
- 2 answers
Posted by Aman Ahammad K 6 years, 10 months ago
- 1 answers
Ram Kushwah 6 years, 10 months ago
{tex}\begin{array}{l}\left(\frac{1-\mathrm{tanA}}{1-\mathrm{cotA}}\right)^2=\left(\frac{1-\mathrm{tanA}}{1-{\displaystyle\frac1{\mathrm{tanA}}}}\right)^2\\=\left(\mathrm{tanA}\frac{(1-\mathrm{tanA})}{\mathrm{tanA}-1}\right)^2\\=(-\mathrm{tanA})^2=\tan^2\mathrm A\end{array}{/tex}
Posted by Rahul Singh 6 years, 10 months ago
- 3 answers
Ram Kushwah 6 years, 10 months ago
Distance traveled in one side of middle flag
=2x(2+4+6,------------+26)
=2x2(1+2+3+...............+13)
=4x13*14/2= 364 m
total distance
left + right
= 2 x 364 = 728 m
Posted by Samyukta ... 6 years, 10 months ago
- 3 answers
Fatma Naayab 6 years, 10 months ago
Posted by Parman Chahal 6 years, 10 months ago
- 1 answers
Shabnam Khatun Naseem Ahamed Siddiqui 6 years, 10 months ago
Posted by Mili Mishra 6 years, 10 months ago
- 1 answers
Posted by Sindhu Lekshmy 6 years, 10 months ago
- 0 answers
Posted by Vineeth Reddy 6 years, 10 months ago
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Posted by Sahil Rajput 6 years, 10 months ago
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Posted by Rohit Prajapati 6 years, 10 months ago
- 2 answers
Ram Kushwah 6 years, 10 months ago
a=5,d=8 l=121
an=a+(n-1)d
121=5+(n-1)x8
116=8n-8
8n=124
n=124/8=15.5
So 121 cannot be term of this AP,So question is not correct(this is not an AP)
pl check question again
Posted by Chirag Sharma 6 years, 10 months ago
- 1 answers
Posted by Nikhil Kumar 6 years, 10 months ago
- 1 answers
Posted by Devansh Aggarwal 6 years, 10 months ago
- 3 answers
Posted by Jio Sim 6 years, 10 months ago
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Posted by Jio Sim 6 years, 10 months ago
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Posted by Rutik Ohal 6 years, 10 months ago
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Posted by Deepak Shukla 6 years, 10 months ago
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Posted by Mohit Rao 6 years, 10 months ago
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Posted by Harjot Singh 6 years, 10 months ago
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Deepanshu Mahajan 6 years, 10 months ago
Posted by Keerthi . 6 years, 10 months ago
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Dpsc Singh 6 years, 10 months ago
Posted by Ranveer Singh 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago

Let O be the position of the bird and B be the position of the boy. Let FG be the building and G be the position of the girl.
In {tex}\triangle{/tex}OLB,
{tex}\frac { OL } { B O }= \sin 30 ^ { \circ }{/tex}
{tex}\Rightarrow \quad \frac { O L } { 100 } = \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow{/tex} OL = 50 m
OM = OL - ML
= OL - FG
= 50 - 20 = 30 m
In {tex}\triangle{/tex}OMG
{tex}\frac {OM}{OG}=sin45°=\frac{1}{√2}{/tex}
OG = OM√2 = 30 √2 = 42.3 meter
Posted by Saksham Kapoor 6 years, 10 months ago
- 2 answers
Devanshu Sharma 6 years, 10 months ago
Posted by Deepali Rana 6 years, 10 months ago
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Posted by Angel Vishakha 6 years, 10 months ago
- 2 answers
Fatma Naayab 6 years, 10 months ago
Devanshu Sharma 6 years, 10 months ago
Posted by Angel Vishakha 6 years, 10 months ago
- 1 answers

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Ayush Shukla???? 6 years, 10 months ago
0Thank You