Let BE = 22 m be the height of the tower and AD = 1.5m be the height of the observer. The point D be the observer's eye. Draw DC \\ AB.
Then, AB = 20.5 m = DC and, EC = BE - BC = BE - AD = 22 -1.5 = 20.5 m [∵ BC = AD]
Let S be the angle of elevation make by observer's eye to the top of the tower i.e. ∠EDC = θ.

In right angled ΔDCE,
\(\tan \theta=\frac{P}{B}=\frac{C E}{D C}=\frac{20.5}{20.5}=1\)
45\(\unicode{xb0} \)

100°

Here, Number=12ⁿ where n stand for any natural number .
Now 12ⁿ= (2² x3)ⁿ
Now , For 12ⁿ to end with 0, it should have 2 as well as 5 in its Prime factors to end with 0, Also to end with 5 , it requires at least a single multiple of 5 in its Prime Factors, So 12ⁿ cannot end with the digit 0 or 5.

1

Yes.

No

x³ + 6x² + 3x + 10 thus we know that,
a + a + b + a + 2b = 6
\(\Rightarrow\) 3(a + b) = 6 \(\Rightarrow\) a + b = 2
a(a + b) + (a + b)(a + 2b) + a(a + 2b) = 3
\(\Rightarrow\) (a + b)(2a + 2b) + a(a + 2b) = 3
\(\Rightarrow\) 2 x 2 x 2 + a² + 2ba = 3
 \(\Rightarrow\)  a² +2ba=-5 \(\Rightarrow\) ab= \(\frac{-5-a^{2}}{2}\)
a(a+b)(a+2b) = -1
a(2)(a+b) = -1
\( a^{2}+a b =-\frac{1}{2} \) 
\(\Rightarrow \quad a^{2}+\left(\frac{-5-a^{2}}{2}\right) =-\frac{1}{2}\)
\(\Rightarrow\) 2a² - 5 - a² = - 1 ~ a2 = 4
 \(\Rightarrow\) a= ± 2
 a+b=2
\(\Rightarrow\)  b=0 or 4

Let the two numbers be x and y, then \(\frac { x }{ y } =\frac { 5 }{ 6 } \quad \Rightarrow \quad y=\frac { 6x }{ 5 } \quad ..(i)\)
Also, \(\frac { x-8 }{ y-8 } =\frac { 4 }{ 5 } \quad \Rightarrow \quad 5x-4y=8\quad \quad ..(ii)\)
Draw graphs of Eq. (i) and Eq. (ii) to get the required numbers.
Numbe0 are 49 and 48.

Here, 327.7081 is terminating. So, it represents a rational number.
Thus, \(327.7081=\frac { 3277081 }{ 10000 } =\frac { p }{ q } \)
Here, q = 10⁴ = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
= 2⁴ x 5⁴ = (2 x 5)⁴
So, the prime factors of q are 2 and 5.

Given, triangle ABC is an equilateral triangle.
\(\therefore \quad \angle A=\angle B=\angle C={ 60 }^{ 0 }\quad and\quad radius,\quad r=\frac { 10 }{ 2 } cm=5cm\)
Area of sector AFEA\(=\frac { \theta }{ { 360 }^{ 0 } } \times \pi { r }^{ 2 }\)
\(=\frac { { 60 }^{ 0 } }{ { 360 }^{ 0 } } \times \pi { \times (5) }^{ 2 }\)
\(=\frac { 25 }{ 6 } \pi \quad { cm }^{ 2 }\)
Since, area of all three sectors are equal.
Total area of shaded region
=3 x Area of sector AFEA\(=3\left( \frac { 25 }{ 6 } \pi \right) \)
\(=3\times \frac { 25 }{ 6 } \times 3.14=39.25\quad { cm }^{ 2 }\)
Hence, the area of shaded region is 39.25 cm²

It is clear from the figure that,
Perimeter of \(\Delta\)AOB
=Distance (AO) + Distance (OB) + Distance (AB)

\(=\sqrt { { (0-0) }^{ 2 }+{ (4-0) }^{ 2 } } +\sqrt { { (0-3) }^{ 2 }+{ (0-0) }^{ 2 } } +\sqrt { { (0-3) }^{ 2 }+{ (4-0) }^{ 2 } } \)
[ using distance formula]
\(=\sqrt { { ({ x }_{ 1 }-{ x }_{ 2 }) }^{ 2 }+{ ({ y }_{ 1 }-{ y }_{ 2 }) }^{ 2 } } \)
\(=\sqrt { { (4) }^{ 2 } } +\sqrt { { (3) }^{ 2 } } +\sqrt { 9+16 } \)
= 4+3+\(\sqrt { 25 } \) = 4 + 3 + 5 = 12 units
Hence, the perimeter of given triangle is 12 units.

Hint : First find the type of triangle using distance formula and hence obtain the angles.
Ans. 60°, equilateral triangle.

33 cm

Here, radius of the hemispherical part = r (say)
Let the total height of the building be h
And the height of the cylindrical part be H
\(\because\)  [ Base of diameter of the done ] = \({{2}\over{3}}\) [ Total height of the building ]
\(\therefore\)  2r = \({{2}\over{3}}h\)
\(\Rightarrow\)   r = \({{1}\over{2}}\times{{2}\over{3}}h={{h}\over{3}},\)
\(\therefore\)   H = \(h - {{h}\over{3}}={{2}\over{3}}h\) meters

Now, Volume of air inside the building = Volume of air inside the done + Volume of air inside the cylindrical part
\(={{2}\over{3}}{\pi r}^{3}+{\pi r}^{2}H\)
\(={{2}\over{3}}\pi\left[ {{h}\over{3}} \right]^{3}+\pi\left[ {{h}\over{3}} \right]^{2}\times\left[ {{2}\over{3}}h \right]\)
\(={\pi}\times{{8}\over{81}}\times{h}^{3}{m}^{3}\)
But, volume of the air in the building
\(={67}{{1}\over{27}}\) m³
\(\therefore\)   \({\pi}\times{{8}\over{81}}\times{h}^{3}=67{{1}\over{21}}\) m³
\(\Rightarrow\)  \({{22}\over{7}}\times{{8}\over{81}}\times{h}^{3}={{1408}\over{21}}\)
\(\Rightarrow\)  h³ = \({{1408}\over{21}}\times{{7}\over{22}}\times{{81}\over{8}}\)
\(\Rightarrow\)  h³ = 8 X 27 = 216
\(\Rightarrow\)  h³ = (6)³
\(\Rightarrow\)   h = 6 m
Hence, the required height of the building is 6 meters.

Since PQ is parallel to the tangent drawn at the point R and radius OR is perpendicular to the tangent. 
ஃ OR ⊥ PQ 
ஃ OL bisects the chord PQ. 
ஃ   PL = LQ 
ஃ  arc PR arc RQ 
i.e., R bisects arc PRQ

Let the three parts of 207 be a -d, and a  + d
\(\therefore\)     a - d + a + a + d = 207
3a  = 207
a = \(\frac{207}{3}\)
a = 69
Also,   ( a - d ) a = 4623
\(\Rightarrow\)    ( 69 - d ) 69 = 4623
\(\Rightarrow\)      69 - d = 67
\(\Rightarrow\)  d = 69 - 67 = 2
Hence, the three parts of 207are 69 - 2, 69 + 2
i.e., 67, 69 and 71.

Total number of enevelopes = 1000
Let A = envelope contains no cash
Number of envelopes containing no cash
 = 1000 - (10 + 100 + 200) = 690
\(\therefore\) P(A) = \(\frac{690}{100}=\frac{69}{100}=0.69\)

\(m=\frac { 19 }{ 14 } \)

scalene triangle

\(45^{\circ}\)

n=5, 11 here both answers are correct be cause sum of 6th term to 11trm is zero

3

Yes

Yes

Here,   t₈ = \(\frac{{t}_{2}}{2}\Rightarrow\) a + 7d = \(\frac{a+d}{2}\)
\(\Rightarrow\)  2a + 14d = a + d \(\Rightarrow\) a = - 13d (i)
and  t₁₁-\(\frac{{t}_{4}}{3}=1\)
\(\Rightarrow\)  a + 10d - \(\frac{a+3d}{3}=1\)
\(\Rightarrow\frac{3a+30d-a-3d}{3}=1\)
\(\Rightarrow\)  2a + 27d = 3
\(\Rightarrow\)   2 ( - 13d ) + 27d = 3  [ using (i) ]
\(\Rightarrow\)     -26d + 27d = 3
\(\Rightarrow\)   d = 3,
\(\therefore\)  from (i)  a = -13 x 3 = - 39
Therefore,,  t₁₅ = a + 14d
= - 39 + 14 x 3
= - 39 + 42 = 3