In Figure  AB denotes the height of the building, BD the flagstaff and P the given point. Note that there are two right triangles PAB and PAD. We are required to find the length of the flagstaff, i.e., DB and the distance of the building from the point P, i.e., PA.
Since, we know the height of the building AB, we will first consider the right \(\Delta\) PAB.
We have \(\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AP}}\) 
i.e., \(\frac{1}{\sqrt{3}}=\frac{10}{\mathrm{AP}}\) 
Therefore, \(\mathrm{AP}=10 \sqrt{3}\)
i.e., the distance of the building from P is 10 \(\sqrt3\) m = 17.32 m.
Next, let us suppose DB = x m. Then AD = (10 + x) m.
Now, in right Δ PAD, \(\tan 45^{\circ}=\frac{\mathrm{AD}}{\mathrm{AP}}=\frac{10+x}{10 \sqrt{3}}\)
Therefore, \(1=\frac{10+x}{10 \sqrt{3}}\) 
i.e., \(x=10(\sqrt{3}-1)=7.32\) 
So, the length of the flagstaff is 7.32 m.

 cos A cos C − sin A sin C
\(=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0\)

LHS = \(\frac { \cos { A } -\sin { A } +1 }{ \cos { A } +\sin { A } -1 } \)
On dividing numerator and denominator by sin A, we get
\(=\frac { \frac { \cos { A } }{ \sin { A } } -\frac { \sin { A } }{ \sin { A } } +\frac { 1 }{ \sin { A } } }{ \frac { \cos { A } }{ \sin { A } } +\frac { \sin { A } }{ \sin { A } } -\frac { 1 }{ \sin { A } } } =\frac { \cot { A } -1+cosecA }{ \cot { A } +1-1cosecA } \) \(\left[ \because \cot { A } =\frac { \cos { \theta } }{ \sin { \theta } } and\frac { 1 }{ \sin { \theta } } =cosec \theta \right] \)
\(=\frac { \cot { A } +cosecA-1 }{ \cot { A } +1-1cosecA } \)
\(=\frac { (\cot { A } +cosecA)-({ cosec }^{ 2 }A-\cot ^{ 2 }{ A } ) }{ \cot { A } +1-1cosecA } \left[ \because 1={ cosec }^{ 2 }A-\cot ^{ 2 }{ A } \right] \)
\(=\frac { (\cot { A } +cosecA)-\left[ (\cot { A } +cosecA)(\cot { A } -cosecA) \right] }{ \cot { A } +1-1cosecA } \) \(\left[ \because \quad { a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b) \right] \)
\(=\frac { (\cot { A } +cosecA)-\left[ 1-(cosecA-\cot { A } ) \right] }{ \cot { A } +1-1cosecA } \)
\(\quad =\frac { (\cot { A } +cosecA)-\left[ 1-cosecA+\cot { A } \right] }{ \cot { A } +1-1cosecA } \)
\(=cosecA+\cot { A } =RHS\)
Hence proved.

Let AB=h m be the height of the tower. Let C is point on ground and AD=5 m be height of flag staff .
Let BC=x m, ㄥACB=45^o and ㄥDCB=60^o
Consider rt . angled ∆ABC, we have
\(\frac { AB }{ BC } \)=tan45⁰
⇒ \(\frac { h }{ x } \)=1   ⇒ h=x
\(\frac { BD }{ BC } \)=tan60^o
⇒ \(\frac { h+5 }{ x } =\sqrt { 3 } \)
⇒ h+5=\(\sqrt { 3 } \)
⇒ h+5=\(\sqrt { 3 } \)h   [∵ from (i), h=x]
5=\(\sqrt { 3 } \)h-h
⇒ h(\(\sqrt { 3 } \)-1)=5
⇒ h=\(\frac { 5 }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } \)
=\(\frac { 5(\sqrt { 3 } +1) }{ 3-1 } =\frac { 5(1.732+1) }{ 2 } \)
=\(\frac { 5\times 2.732 }{ 2 } \)=5 x 1.366
=6.83

Let BC be the building, AB be the transmission tower and D be the point on the ground from where the angles of elevations are to be measured.

\(\begin{array}{rlrl} \text { In } \triangle B C D, & \tan 45^{\circ} =\frac{B C}{C D} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow 1 =\frac{20}{C D} \Rightarrow C D=20 \mathrm{~m} \end{array}\)
\(\begin{array}{llrl} \text { In } \triangle A C D, & \tan 60^{\circ} =\frac{A C}{C D} \end{array}\)
\(\begin{array}{llrl} \Rightarrow \sqrt{3} =\frac{A B+B C}{C D} \\ \end{array}\)
\(\begin{array}{llrl} \Rightarrow \sqrt{3} =\frac{A B+20}{20} \\ \end{array}\)
\(\Rightarrow A B =20 \sqrt{3}-20=20(\sqrt{3}-1) \mathrm{m}\)
Thus, the height of the tower is \(20(\sqrt{3}-1) \mathrm{m}\).

OP = Radius of the circle OA = 5 cm; AP = 4 cm
o [RAdius and tangent are ⊥ ar]
OA² = AP² + OP²   [By pythagoras theorem]
5² = 4² + OP²
⇒ 25 = 16 + OP²  ⇒ 25 - 16 = OP² ⇒ 9 = OP² ⇒ OP = \(\sqrt9\) = 3
Radius = 3 cm

\(\begin{aligned} \mathrm{LHS} & =\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A} \\ \end{aligned}\)
\(\begin{aligned} =\frac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A-\cos A)(\sin A+\cos A)} \end{aligned}\)
\(=\frac{\left[\begin{array}{c} \sin ^2 A+2 \sin A \cos A+\cos ^2 A+\sin ^2 A \\ -2 \sin A \cos A+\cos ^2 A \end{array}\right]}{\sin ^2 A-\cos ^2 A}\)
\(\left[\because(a \pm b)^2=a^2+b^2 \pm 2 a b\right]\)
\(\begin{aligned} & =\frac{2 \sin ^2 A+2 \cos ^2 A}{\sin ^2 A-\cos ^2 A}=\frac{2\left(\sin ^2 A+\cos ^2 A\right)}{\sin ^2 A-\cos ^2 A} \end{aligned}\)
\(\begin{aligned} =\frac{2}{\sin ^2 A-\cos ^2 A} \end{aligned}\)
\(\begin{aligned} =\frac{2}{\sin ^2 A-\left(1-\sin ^2 A\right)} \end{aligned}\)
\(\begin{aligned} =\frac{2}{\sin ^2 A-1+\sin ^2 A} \end{aligned}\)
\(\begin{aligned} =\frac{2}{2 \sin ^2 A-1} \end{aligned}\)
= RHS                                   Hence proved.

In \(\Delta ADC\), tan 30°=\(\frac { H-200 }{ x } \)
\(\frac { 1 }{ \sqrt { 3 } } =\frac { H-200 }{ x } \)
x =\(\sqrt{3}\)(H-200)m
In \(\Delta ADF\), tan 60° =\(\frac { H+200 }{ x } \)
\(\sqrt{3}\)=\(\frac { H+200 }{ x } \)
\(\sqrt{3}\)=\(\frac { H-200 }{ \sqrt { 3 } (H-200) } \)

3(H -200) = H + 200
3H - H = 200 + 600
2H=800
So, required height H = 400 m

To prove \(\left( \cot { \theta } -cosec\theta \right) ^{ 2 }=\frac { 1-\cos { \theta } }{ 1+\cos { \theta } } \)
\(LHS=\left( \cot { \theta } -cosec\theta \right) ^{ 2 }\)
\(=\left( \frac { \cos { \theta } }{ \sin { \theta } } -\frac { 1 }{ \sin { \theta } } \right) ^{ 2 }\)
\(=\left( \frac { \cos { \theta } -1 }{ \sin { \theta } } \right) ^{ 2 }\)
\(=\frac { { \left( 1-\cos { \theta } \right) }^{ 2 } }{ \sin ^{ 2 }{ \theta } } \)    \(\left( \because \quad \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right) \)
\(=\frac { { \left( 1-\cos { \theta } \right) }^{ 2 } }{ \left( 1-\cos ^{ 2 }{ \theta } \right) } \)
\(=\frac { \left( 1-\cos { \theta } \right) \left( 1-\cos { \theta } \right) }{ \left( 1-\cos { \theta } \right) \left( 1+\cos { \theta } \right) } \)
\(\frac { 1-\cos { \theta } }{ 1+\cos { \theta } } \)
= RHS

\(LHS=\frac { cosec\theta +cot\theta }{ cosec\theta -cot\theta } =\frac { \frac { 1 }{ sin\theta } +\frac { cos\theta }{ sin\theta } }{ \frac { 1 }{ sin\theta } -\frac { cos\theta }{ sin\theta } } \)
\(=\frac { (1+cos\theta )/sin\theta }{ (1-cos\theta )/sin\theta } =\frac { 1+cos\theta }{ 1-cos\theta } \)
\(=\frac { 1+cos\theta }{ 1-cos\theta } \times \frac { 1+cos\theta }{ 1+cos\theta } \)
\(=\frac { { \left( 1+cos\theta \right) }^{ 2 } }{ (1-cos\theta )(1+cos\theta ) } =\frac { { \left( 1+cos\theta \right) }^{ 2 } }{ 1-{ cos }^{ 2 }\theta } \)
\(=\frac { 1+{ cos }^{ 2 }\theta +2cos\theta }{ { sin }^{ 2 }\theta } \)
\(=\frac { 1 }{ { sin }^{ 2 }\theta } +\frac { { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta } +\frac { 2cos\theta }{ { sin }^{ 2 }\theta } \)
\(={ cosec }^{ 2 }\theta +{ cot }^{ 2 }\theta +2{ cosec }^{ 2 }\theta cos\theta \)
\(=1+{ cot }^{ 2 }\theta +{ cot }^{ 2 }\theta +2{ cosec }^{ 2 }\theta cos\theta \)
\(=1+2{ cot }^{ 2 }\theta +2{ cosec }^{ 2 }\theta cos\theta =RHS\)

Since we will apply the identity involving sec \(\theta\) and tan \(\theta\), let us first convert the LHS (of the identity we need to prove) in terms of sec \(\theta\) and tan \(\theta\) by dividing numerator and denominator by cos \(\theta\).
\(\begin{aligned} \text { LHS } & =\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1}=\frac{\{(\tan \theta+\sec \theta)-1\}(\tan \theta-\sec \theta)}{\{(\tan \theta-\sec \theta)+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{\left(\tan ^2 \theta-\sec ^2 \theta\right)-(\tan \theta-\sec \theta)}{\{\tan \theta-\sec \theta+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)} \end{aligned}\)
\(=\frac{-1}{\tan \theta-\sec \theta}=\frac{1}{\sec \theta-\tan \theta}\)
which is the RHS of the identity, we are required to prove.

60°

We are given two concentric circles C₁ and C₂ with centre O and a chord AB of the larger circle C₁ which touches the smaller circle C₂ at the point P (see Fig). We need to prove that AP = BP.

Let us join OP. Then, AB is a tangent to C₂ at P and OP is its radius. Therefore, by Theorem.
OP \(\perp\) AB
Now AB is a chord of the circle C₁ and OP \(\perp\) AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord,
i.e., AP = BP

Let ABCD is a quadrilateral circumscribing a circle with centre O. Let circle touches the sides of a quadrilatcral at points E, F, G and H.

To prove \(\angle\)AOB + \(\angle\)COD = 180°
and \(\angle\) AOD + \(\angle\)BOC = 180°
Construction Join OE, OF, OG and OH.
Proof We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.
and       ....(i)
Also, we know that the sum of all angles subtended at a point is 360°.
\(\begin{array}{rlrl} \therefore \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8 & =360^{\circ} \\ \end{array}\)    ...(ii)
\(\begin{array}{rlrl} \Rightarrow 2(\angle 2+\angle 3+\angle 6+\angle 7) & =360^{\circ} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & (\angle 2+\angle 3)+(\angle 6+\angle 7)=180^{\circ} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \angle A O B+\angle C O D=180^{\circ} \end{array}\)
Similarly, we have
\(\begin{aligned} 2(\angle 1+\angle 8+\angle 4+\angle 5) & =360^{\circ} \\ \end{aligned}\)  [from Eq. (i) and (ii)]
\(\begin{aligned} & \Rightarrow(\angle 1+\angle 8)+(\angle 4+\angle 5)=180^{\circ} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \angle A O D+\angle B O C=180^{\circ} \end{aligned}\) Hence proved.

\(L H S=\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}\)
\(=\frac{\cos A\left(\frac{1}{\sin A}-1\right)}{\cos A\left(\frac{1}{\sin A}+1\right)}=\frac{\left(\frac{1}{\sin A}-1\right)}{\left(\frac{1}{\sin A}+1\right)}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\operatorname{RHS}\)

Given, \(\sin { \alpha =\frac { 1 }{ 2 } } \)
then \(3\sin { \alpha } -4\sin ^{ 3 }{ \alpha } =3\times \frac { 1 }{ 2 } -4\times { \left( \frac { 1 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 3 }{ 2 } -\frac { 4 }{ 8 } =1\)  

\(\frac {1}{7}\)

4 cm

Distance covered in 5000 revolutions = 11 km.
Distance covered in 1 revolution
Distance covered in 1 revolution \(={11000\over 5000}m={11\over 5}m\)
Distance covered in 1 revolution = circumference of the wheel
\(\Rightarrow\)\(2\ \pi r ={11\over 5} \Rightarrow 2\times {22\over 7}\times r = {11\over 5}\)
\(\Rightarrow\)  \(r \Rightarrow {11\over 5}\times 7\times {1\over 2\times 22}={7\over 20}m\)
\(\therefore\) Diameter \(=2\times r=2\times {7\over 20}={7\over 10}\times 100 cm=70cm\)

Given, radius of a circle, AO = 10 cm and \(\angle\)AOC = 90°
Area of \(\triangle A O C=\frac{1}{2} \times O A \times O C=\frac{1}{2} \times 10 \times 10=50 \mathrm{~cm}^2\)
\(\begin{aligned} \text { Area of sector } O A E C O & =\frac{\theta}{360^{\circ}} \times \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(10)^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{314}{4}=78.5 \mathrm{~cm}^2 \end{aligned}\)

(i) Area of minor segment AECDA
= Area of sector OAECO - Area of \(\Delta\)AOC
= 78.5 - 50 = 28.5 cm²
(ii) Area of major sector OAFGCO
= Area of circle - Area of sector OAECO
= 3.14 \(\times\)(10)² - 78.5
= 314 - 78.5 = 235.5 cm²

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.
In 5 minutes, minute hand will rotate = 360^@/60xx5 = 30^@
Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.
Area of sector of angle θ = \(\frac{\theta}{360^{\circ}} \times \pi r^{2}\)
Area of sector of 30° \(=\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14\)
\(\begin{array}{l} =\frac{22}{12} \times 2 \times 14 \\ =\frac{11 \times 14}{3} \end{array}\)
=154/3 cm²
Therefore, the area swept by the minute hand in 5 minutes is 154/3 cm²

Let BC = 50 m be the height of the tower and AD = h m be the height of the building. Angle of elevation of the top of the building from the foot of the tower is \(\angle\)DBA = 30° and angle of elevation of the top of the tower from the foot of the building \(\angle\)CAB = 60°. Also, let AB = x m be the distance between foots of the tower and the building.
In right angled \(\Delta\)BAD, tan 30°=\(\frac{P}{B}=\frac{A D}{A B}\)
\(\begin{array}{lll} \Rightarrow & \frac{1}{\sqrt{3}}=\frac{h}{x} & {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]} \\ \end{array}\)
\(\begin{array}{lll} \Rightarrow & b=\frac{x}{\sqrt{3}} \end{array}\)   ...(i)
and in right angled \(\Delta\)CBA, tan 60° = \(\frac{B C}{A B}\)
\(\begin{array}{lll} \Rightarrow & \sqrt{3}=\frac{50}{x} & {\left[\because \tan 60^{\circ}=\sqrt{3}\right]} \\ \end{array}\)
\(\begin{array}{lll} \Rightarrow & x=\frac{50}{\sqrt{3}} \mathrm{~m} \end{array}\)
On putting \(x=\frac{50}{\sqrt{3}}\) in Eq. (i), we get
\(h=\frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}}=\frac{50}{3}=16 \frac{2}{3} \mathrm{~m}\)
Hence, the height of the building is \(16 \frac{2}{3} \mathrm{~m}\).

If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(i) (c): Area of the region containing blue colour
\(=\frac{22}{7} \times 5 \times 5 \times \frac{80^{\circ}}{360^{\circ}}-\frac{22}{7} \times 3 \times 3 \times \frac{80^{\circ}}{360^{\circ}}\)
\(=\frac{22}{7} \times \frac{2}{9} \times[25-9]=\frac{44}{63}(16)=11.17 \mathrm{~cm}^{2}\)
(ii) (b) : Area of the region containing green colour
\(=\frac{22}{7} \times \frac{60^{\circ}}{360^{\circ}}[5 \times 5-3 \times 3]=\frac{22}{7} \times \frac{1}{6} \times 16=8.38 \mathrm{~cm}^{2}\)
(iii) (d): Perimeter of the region containing red colour
= 2 + 2 + length of arc of sector having radius 3 cm + length of arc of sector having radius 5 cm.
\(=4+2 \times \frac{22}{7} \times 3 \times \frac{20^{\circ}}{360^{\circ}}+2 \times \frac{22}{7} \times 5 \times \frac{20^{\circ}}{360^{\circ}}\)
\(=4+\frac{44}{7} \times \frac{1}{18} \times 8=4+\frac{176}{63}=4+2.79=6.79 \mathrm{~cm}\)
(iv) (a): Required area \(=\frac{22}{7} \times 3 \times 3 \times \frac{160^{\circ}}{360^{\circ}}\)
\(=\frac{88}{7}=12.57 \mathrm{~cm}^{2}\)
(v) (a): Angle of given sector = 800 + 600 + 200 = 1600
Thus, the given region represents minor sector of a circle.

(i) (b): 

Here, OS the is radius of circle.
Since radius at the point of contact is perpendicularto tangent.
So, \(\angle\)OSA = 90°
(ii) (d): Since, length of tangents drawn from an external point to a circle are equal.
\(\therefore\) AS=AP,BP=BQ,
CQ = CR and DR = DS
(iii) (a): AP = AS = AD _ DS = AD _ DR (Using (1)
= 11 - 7 = 4 cm
(iv) (b): In quadrilateral OQCR,

\(\angle\)QCR = 60° (Given)
And \(\angle\)OQC = \(\angle\)ORC = 90° [Since, radius at the point of contact is perpendicular to tangent.]
\(\therefore\) \(\angle\)QOR = 360° - 90° - 90° - 60° = 120°
(v) (c): From (1), we have AS = AP, DS = DR,
BQ = BP and CQ = CR
Adding all above equations, we get
AS + DS + BQ + CQ = AP + DR + BP + CR
\(\Rightarrow\) AD + BC = AB + CD

(i) (b): The person who makes small angle of elevation is more closer to the balloon.
\(\therefore\) Radlra is more closer to the balloon.
(ii) (b): \(\text { In } \Delta E F D, \tan 30^{\circ}=\frac{E D}{D F}\)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{D F} \)
\(\Rightarrow \quad D F=h \sqrt{3} \mathrm{~m}\)
(iii) (a): In \(\Delta\)GCE,
\(\begin{array}{l} \tan 60^{\circ}=\frac{E C}{G C}=\frac{h+4}{D F} \\ \Rightarrow \quad \sqrt{3}=\frac{h+4}{\sqrt{3} h} \Rightarrow 3 h=h+4 \Rightarrow h=2 \end{array}\)
(iv) (c): Height of the balloon from the ground = BE = BC + CD + DE = 2 + 4 + 2 = 8 m
(v) (b)