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Krithi ♀️
Subject: Maths
, asked on 9/5/18
value of sin 45 in decimals.
Answer
1
Raghav Singh
Subject: Maths
, asked on 30/3/18
Please answer q 16. I reached the cot ? tan point but can't solve it further. Please explain in brackets side by side.
16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complimentary. Prove that the height of the tower is 6 m.
Please answer q 16. I reached the cot ? tan point but cant solve it further. Please explain in brackets side by side.
Answer
2
Son Ser
Subject: Maths
, asked on 30/3/18
Plzz fast i have exams
Q.
${\left(\mathrm{sin}\theta +sec\theta \right)}^{2}+\left(\mathrm{cos}\theta +\mathrm{cos}e{c}^{2}\theta \right)={\left(1+sec\theta \mathrm{cos}ec\theta \right)}^{2}$
Answer
3
Raghav Singh
Subject: Maths
, asked on 29/3/18
In this i am getting opposite values of tan and cot. I am getting 4cotx & 9 tanx and also how are both of them equal to 36 ? -
19:23 Let AB be the tower. C and D be the two points with distance 4 m and 9 m from the base respectively. A/q, In right
AABC, tan x = AB/BC tan x = AB/4 AB = 4 tan x
(i) also, In right AABD, tan (900-x) = AB/BD cot x = AB/9 AB = 9 cot x
(ii) Multiplying eqn (i) and (ii) AB2 = 9 cot xx 4 tan x AB2 = 36
Height cannot be negative. Therefore, the height of the tower is 6 m. Hen Proved.
Answer
1
Mohit Sharma
Subject: Maths
, asked on 27/3/18
Problems of height and distance can be solved by more than one way . Is full marks are awarded for the correct answer of problem done in different approach which is in syllabus in boards ? Pls tell trm is my papaer
Answer
2
Champ Jee
Subject: Maths
, asked on 27/3/18
Solve this :
Q. An aeroplane, when flying at a height of 4000 m from the ground passes vertically above another
aeroplane at an instant when the angles of elevation of the two planes from the same point on the
ground are 60
^{o}
and 45
^{o}
respectively. Find the vertical distance between the aeroplanes at that instant.
$\left(\mathrm{Take}\sqrt{3}=1.73\right)$
Answer
3
Anu
Subject: Maths
, asked on 26/3/18
Dear experts plzz clarify!!!!!
Answer
2
Riya
Subject: Maths
, asked on 26/3/18
Please do qs no. 27(i)
Q.27.(i)
$\frac{1+\mathrm{cos}\theta +\mathrm{sin}\theta}{1+\mathrm{cos}\theta -\mathrm{sin}\theta}=\frac{1+\mathrm{sin}\theta}{\mathrm{cos}{\displaystyle \theta}}$
Answer
1
Raghav Singh
Subject: Maths
, asked on 26/3/18
In this q what have we dine to PD in the last step. Is it rationalisation ?
Answer
1
Akshit Gupta
Subject: Maths
, asked on 26/3/18
Solve this:
Q.61. From the top of a 10 m high building, the angle of depression of the foot of a tower is 30
$\xb0$
and the angle of elevation of top of the tower from the foot of the building is 60
$\xb0$
. Find the distance between the tops of the building and the tower.
[Use
$\sqrt{7}$
= 2.64]
Answer
1
अर्कज ..
Subject: Maths
, asked on 25/3/18
Q.15 A ladder against a vertical wall at an inclination
$\alpha $
to the horizontal .The foot is pulled away from the wall through a distance p, so that its upper end slides a distance q down the wall and then the ladder makes an angle
$\beta $
to the horizontal . show that p /q =
$\mathrm{cos}\beta -\mathrm{cos}\alpha /\mathrm{sin}\alpha -\mathrm{sin}\beta $
Answer
1
Murugan
Subject: Maths
, asked on 25/3/18
A round balloon of radious 'a' ,subtends an angle Theeta at the eye of a observer while the angle of elevation of its centre is phi ..prove that the height of the centre of the balloon is = sin phi cosec theeta +2...
Answer
2
Khushi Jain
Subject: Maths
, asked on 25/3/18
The ratio of the height of a tower and the length of its shadow on the ground is
$\sqrt{3}:1$
. What is the angle of elevation of the sun?
Answer
1
Prateek Kumar
Subject: Maths
, asked on 24/3/18
All questions please
Two stations due south of a leaning tower which leans towards the north are at distances a and b from its foot. If
$\alpha and\beta $
be the elevations of the top of the tower from these stations, its inclination
$\theta $
to the horizontal is given by cot
$\theta $
is equal to
$\left(A\right)\frac{bcot\alpha -acot\beta}{b-a}\phantom{\rule{0ex}{0ex}}\left(B\right)\frac{bcot\alpha +acot\beta}{b+a}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(C\right)\frac{acot\alpha -bcot\beta}{a-b}\phantom{\rule{0ex}{0ex}}\left(D\right)\frac{bcot\beta -acot\alpha}{a+b}$
Answer
1
Ibtihal ..
Subject: Maths
, asked on 23/3/18
Solve this:
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is
$60\xb0$
and the angle of depression of the foot of the tower is
$30\xb0$
. Find the height of the tower.
Answer
1
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What are you looking for?

16.The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complimentary. Prove that the height of the tower is 6 m.Please answer q 16. I reached the cot ? tan point but cant solve it further. Please explain in brackets side by side.

Q. ${\left(\mathrm{sin}\theta +sec\theta \right)}^{2}+\left(\mathrm{cos}\theta +\mathrm{cos}e{c}^{2}\theta \right)={\left(1+sec\theta \mathrm{cos}ec\theta \right)}^{2}$

19:23 Let AB be the tower. C and D be the two points with distance 4 m and 9 m from the base respectively. A/q, In right

AABC, tan x = AB/BC tan x = AB/4 AB = 4 tan x

(i) also, In right AABD, tan (900-x) = AB/BD cot x = AB/9 AB = 9 cot x

(ii) Multiplying eqn (i) and (ii) AB2 = 9 cot xx 4 tan x AB2 = 36

Height cannot be negative. Therefore, the height of the tower is 6 m. Hen Proved.

Solve this :Q. An aeroplane, when flying at a height of 4000 m from the ground passes vertically above another

aeroplane at an instant when the angles of elevation of the two planes from the same point on the

ground are 60

^{o}and 45^{o}respectively. Find the vertical distance between the aeroplanes at that instant.$\left(\mathrm{Take}\sqrt{3}=1.73\right)$

Q.27.(i) $\frac{1+\mathrm{cos}\theta +\mathrm{sin}\theta}{1+\mathrm{cos}\theta -\mathrm{sin}\theta}=\frac{1+\mathrm{sin}\theta}{\mathrm{cos}{\displaystyle \theta}}$

Q.61. From the top of a 10 m high building, the angle of depression of the foot of a tower is 30$\xb0$ and the angle of elevation of top of the tower from the foot of the building is 60$\xb0$. Find the distance between the tops of the building and the tower.

[Use $\sqrt{7}$ = 2.64]

Two stations due south of a leaning tower which leans towards the north are at distances a and b from its foot. If $\alpha and\beta $ be the elevations of the top of the tower from these stations, its inclination $\theta $ to the horizontal is given by cot $\theta $ is equal to

$\left(A\right)\frac{bcot\alpha -acot\beta}{b-a}\phantom{\rule{0ex}{0ex}}\left(B\right)\frac{bcot\alpha +acot\beta}{b+a}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(C\right)\frac{acot\alpha -bcot\beta}{a-b}\phantom{\rule{0ex}{0ex}}\left(D\right)\frac{bcot\beta -acot\alpha}{a+b}$

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $60\xb0$ and the angle of depression of the foot of the tower is $30\xb0$. Find the height of the tower.