(i) In right angled \(\triangle A B C\),
\(\tan 45^{\circ} =\frac{A B}{B C}\)
\(\Rightarrow  B C  =A B=80 \mathrm{~m}\)
\( {\left[\because A B=80 \mathrm{~m} \text { and } \tan 45^{\circ}=1\right] }\)
Therefore Required distance \(=80 \mathrm{~m}\)
(ii) In right angled \(\triangle D E C\),
\(\tan 30^{\circ} =\frac{D E}{E C}\)
\(\Rightarrow  \frac{1}{\sqrt{3}}  =\frac{80}{E C} \)
\(\Rightarrow \left[\because D E=80 \mathrm{~m} \text { and } \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]\)
\(\Rightarrow \quad E C =80 \sqrt{3} \mathrm{~m}\)
\(\therefore \quad B E  =E C-B C\)
\(=80 \sqrt{3}-80 \quad \text { [using Eq. (i)] }\)
\(=80(\sqrt{3}-1) \mathrm{m} \)
Hence, distance the bird flew
\(=A D=B E =80(\sqrt{3}-1) \mathrm{m}\)
Or 
In right angled \(\triangle C G F\),
\( \tan 60^{\circ}  =\frac{G F}{G C} \)
\(\Rightarrow  \sqrt{3}  =\frac{80}{G C}\)
\( {\left[\because G F=E D=80 \mathrm{~m} \text { and } \tan 60^{\circ}=\sqrt{3}\right]} \)
\(\Rightarrow  G C  =\frac{80 \sqrt{3}}{3} \mathrm{~m}\)
 

Distance travelled by the ball after hitting the tree =AF = BG
and BG =BC-GC 
\(=\left(80-\frac{80 \sqrt{3}}{3}\right)\)
\(=80\left(\frac{3-\sqrt{3}}{3}\right) \mathrm{m}\) [using Eqs. (i) and (ii)]
(iii) \( \text { Speed of the bird }=\frac{\text { Distance }}{\text { Time }} \)
\(=\frac{20(\sqrt{3}+1)}{2} \mathrm{~m} / \mathrm{s}\)
\(=10(\sqrt{3}+1) \mathrm{m} / \mathrm{s}\)
\(=600(\sqrt{3}+1) \mathrm{m} / \mathrm{min} \)
\(\left[\because 1 \mathrm{~s}=\frac{1}{60} \mathrm{~min}\right]\)

(a) Length of the square shaped grass field = 20 m

Area of the square shaped grass field = 20 \(\times\) 20 = 400 m²
(b) (i) Length of the rope = 7 m
Thus, each horse can graze upto 7 m of distance along the side.
The grazed area is making a complete circle by taking all the four grazed parts.
So, area of grazed part = \(\pi\)r²
\(=\frac{22}{7} \times 7^2=22 \times 7=154 \mathrm{~m}^2\)
Therefore, area of the total field in which these horses can graze is 154 m².
(ii) New length of the rope of each horse = 10 m
Area grazed by one horse \(=\frac{1}{4} \pi r^2\)
\(\begin{aligned} & =\frac{1}{4} \times 3.14 \times 10^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{314}{4}=78.5 \mathrm{~m}^2 \end{aligned}\)
Therefore, the required area grazed by one horse is 78.5 m².
(c) Length of rope of each horse = 7 m
Area of square shaped grass field = 400 m²
                                                        [from part (a)]
Area of total field grazed by horses = 154 m²
                                                        [from Eq. (ii) (a)]
Area of field left ungrazed = Area of square field - Area of grazed field by horses
= 400 - 154 = 246
Therefore, area of field left ungrazed is 246m².

(i) (c) In quadrilateral PQOR, we have

\(\begin{array}{r} \angle Q P R+\angle P R O+\angle P Q O+\angle R O Q=360^{\circ} \\ \end{array}\)
\(\Rightarrow \begin{array}{r} 30^{\circ}+90^{\circ}+90^{\circ}+\angle R O Q=360^{\circ} \end{array}\)
[\(\because\) radius is always perpendicular to the tangent at point of contact]
\(\Rightarrow \quad \angle R O Q=360^{\circ}-210^{\circ}=150^{\circ} \)
\((ii) (a) In \quad \triangle Q O R, O Q=O R\)     [radii]
\(\begin{array}{ll} \therefore & \angle O R Q=\angle O Q R \\ \end{array}\)
\(\begin{array}{ll} \text { Now, } & \angle R O Q+\angle O R Q+\angle O Q R=180^{\circ} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & 2 \angle O Q R=180^{\circ}-150^{\circ} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & 2 \angle O Q R=30^{\circ} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \angle O Q R=15^{\circ} \end{array}\)
Again, \(\begin{array}{ll} \angle O Q R=90^{\circ} \end{array}\)    \([\because O Q \perp Q P]\)
\(\begin{aligned} \Rightarrow \angle O Q R+\angle R Q P=90 \\ \end{aligned}\)^\({\circ}\)
\(\Rightarrow \quad \angle R Q P=90^{\circ}-15^{\circ}=75^{\circ}\)
(iii) (b) We know that angle subtended by an are at centre is double the angle subtended by it at any other part of the circle.
\(\begin{aligned} 2 \angle R S Q & =\angle R O Q \\ \end{aligned}\)
\(\begin{aligned} \angle R S Q & =\frac{1}{2} \times 150^{\circ}=75^{\circ} \end{aligned}\)
(iv) (a) \(\angle\)ORP = 90^\(\circ\) as radius is always perpendicular to the tangent at the point of contact.

(i) (b) Let AB be the monument of height 42 m and C is the point where they are standing such that BC = 42 m.


Now, in \(\Delta\)ABC,
\(\begin{aligned} & \tan \theta=\frac{A B}{B C} \Rightarrow \tan \theta=\frac{42}{42}=1 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \tan \theta & =1 \quad \Rightarrow \quad \theta=45^{\circ} \end{aligned}\)
(ii) (a) In \(\Delta\)ABC,

\(\begin{aligned} \tan 60^{\circ} & =\frac{A B}{B C} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \sqrt{3} & =\frac{42}{B C} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad B C & =\frac{42}{\sqrt{3}}=\frac{42 \sqrt{3}}{3}=14 \sqrt{3} \\ \end{aligned}\)
\(\begin{aligned} =14 \times 1.732=24.248=24.25 \mathrm{~m} \end{aligned}\)
(iii) (a) Let AB be the height of the tower.

Then, in \(\Delta\)ABC, tan 60°\(=\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{h}{20}\)
\(\Rightarrow\)  h = 20\(\sqrt3\) m
(b) Let h and x be the height and length of shadow of the vertical tower.

Then, in \(\Delta\)ABC,
\(\begin{array}{rlrl} \tan \theta & =\frac{A B}{B C} \Rightarrow \tan \theta=\frac{h}{x} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & & \tan \theta & =1 \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \theta & =45^{\circ} & {[\because h: x=1: 1]} \end{array}\)
(v) (c) The angle of depression of an object viewed, is the angle formed by the line of sight with the horizontal, when it is below the horizontal level.

(a) (iv) NS and NR are both tangent to the circle
So, \(N S \perp S A\) and \(N R \perp R A\) 
[ \(\therefore\)  Tangent to a circle is perpendicular to the radius through the point of contact ]
\(\therefore \angle N S A=90^{\circ}\) 
(b) (ii) 
\(\angle N S A=90^{\circ} \quad \Rightarrow N A^{2}=N S^{2}+S A^{2}\) [ By Pythagora's Theorem]
\(\Rightarrow N S=\sqrt{N A^{2}-S A^{2}}=\sqrt{13^{2}-5^{2}}=\sqrt{169-25}\) 
\(=\sqrt{144}=12 \mathrm{~m}\) 
(c) (iiii)
NR = NS = 12m
Tangents drawn from an external point are equal
\(\therefore N R=12 \mathrm{~m}\) 
(d) (iv)
\(\angle S N R+\angle R A S=180^{\circ} \Rightarrow \angle R A S=180^{\circ}-\angle S N R\) 
[\(\because\) The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact to the centre]
\(\Rightarrow \angle R A S=180^{\circ}-\theta\) 
(e) (i) 
\(\angle R A S=180^{\circ}-\theta\) [ Calulated bove]
Now, \(\angle N A S=\angle N A R=\frac{1}{2} \angle R A S\) 
[\(\because\) If two tangents are drawn from an external point, then they extend equal angles at the centre]
\(\Rightarrow \angle N A S=\frac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\frac{\theta}{2}\)

(i) (a)  \(\cos \theta=0.5=\cos 60^{0}\) 
\(\Rightarrow \theta=60^{\circ}\) 
(ii) (c) BD = AD - AB
= (5-1.3)m
= 3.7m
(iii) (d) \(\frac{\mathrm{BD}}{\mathrm{BC}}=\sin 60^{\circ}\) or \(\frac{3.7}{\mathrm{BC}}=\frac{\sqrt{3}}{2}\) 
\(\mathrm{BC}=\frac{3.7 \times 2}{\sqrt{3}}=4.28 \mathrm{~m}(\text { approx. })\) 
(iv) (b) \(\frac{\mathrm{DC}}{\mathrm{BD}}=\cot 60^{\circ}=\frac{1}{\sqrt{3}}\) 
\(\mathrm{DC}=\frac{3.7}{\sqrt{3}}=2.14 \mathrm{~m}(\text { approx. })\) 
(v) (c) \(\frac{\mathrm{BD}}{\mathrm{BC}}=\sin 60^{\circ}\) or \(\frac{7.4}{\mathrm{BC}}=\frac{\sqrt{3}}{2}\) 
\(\mathrm{BC}=\frac{7.4 \times 2}{\sqrt{3}}=8.56 \mathrm{~m}(\text { approx. })\)

(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) (a): Area of sector ODCO \(=\frac{1}{4} \pi r^{2}\)
\(=\frac{1}{4} \times \frac{22}{7} \times 14 \times 14=154 \mathrm{~cm}^{2}\)
(ii) (b): Area of \(\triangle A O B=\frac{1}{2} \times O A \times O B=\frac{1}{2}(20 \times 20)\)
= 200 cm²
(iii) (a) : Area of region which is golden plated
= area of \(\Delta\)OAB - area of sector ODCO.
= 200 - 154 = 46 cm²
\(\therefore\) Total cost of golden plating = Rs (6 x 46) = Rs 276
(iv) (c): Area of major sector = area of circle - area of minor sector
\(=\pi r^{2}-\frac{1}{4} \pi r^{2}=\frac{3 \pi r^{2}}{4}=\frac{3}{4} \times \frac{22}{7} \times 14 \times 14=462 \mathrm{~cm}^{2}\)
(v) (d): Length of arc DC \(=\frac{90^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 14\)
= 22cm

(i) (a): In a week, Kartik drives his motorbike 3 days to go to college
\(\therefore\)Total distance travelled by Kartik through
motorbike = 2 x 4.2 x 6 = 50.4 km
(ii) (b): In a week Kartik rides his bicycle 3 days to go to college.
\(\therefore\) Total distance travelled by Kartik through bicycle
= Length of arc \(\widehat{A C B} \times 6\)
\(=\frac{\theta}{360^{\circ}} \times 2 \pi r \times 6 \Rightarrow \frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 4.2 \times 6=26.4 \mathrm{~km}\)
(iii) (c) : Area of sector \(A O B=\frac{\theta}{360^{\circ}} \times \pi r^{2}\)
\(=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(4.2)^{2}=9.24 \mathrm{~km}^{2}\)
(iv) (a): Total cost of fuel used for a week = Rs (20 x 50.4) = Rs 1008
(v) (d): Total length of available paths
\(=4.2+4.2+\frac{\theta}{360^{\circ}} \times 2 \pi r \)
\(=8.4+\frac{90^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 4.2=8.4+6.6=15 \mathrm{~km}\)

(i) (b): Let the radius of outer most circle be R.
Outer most circumference = 308 m [Given]
\(\Rightarrow 2 \pi R=308 \Rightarrow 2 \times \frac{22}{7} \times R=308\)
\(\Rightarrow \quad R=\frac{308 \times 7}{2 \times 22}=49 \mathrm{~m}\)
(ii) (c): Let the radius of inner most circle be r
Inner most circumference = 264 m [Given]
\(\Rightarrow 2 \pi r=264 \)
\(\Rightarrow \quad 2 \times \frac{22}{7} \times r=264 \Rightarrow r=\frac{264 \times 7}{2 \times 22}=42 \mathrm{~m}\)
(iii) (a): Width of the track = Radius of outer most track - Radius of inner most track = 49 - 42 = 7 m
(iv) (d): Area of the race track = Area of outer circle - Area of inner circle
\(=\pi\left(R^{2}-r^{2}\right)=\pi\left[(49)^{2}-(42)^{2}\right] \)
\(v=\frac{22}{7}[2401-1764]=2002 \mathrm{~m}^{2}\)
(v) (b): Cost of painting the whole race track
= Rs (6 x 2002) = Rs 12012.

(i) (b): Since \(\Delta\)ABC is a right angled triangle
\(\therefore\)  Area of triangle ABC \(=\frac{1}{2} \times 7 \times 24=84 \mathrm{~m}^{2}\)
(ii) (a): Area of region grazed by the cow
\(=\frac{\angle A}{360^{\circ}} \times \pi \times(3.5)^{2}\)
(iii) (d): Area of the region grazed by the buffalo and the horse
\(=\frac{\angle B}{360^{\circ}} \times \pi \times(3.5)^{2}+\frac{\angle C}{360^{\circ}} \times \pi \times(3.5)^{2} \)
\(=\frac{(\angle B+\angle C)}{360^{\circ}} \times \pi \times(3.5)^{2}\)
(iv) (d): Total area grazed by the cow, the horse and the buffalo
\(=\frac{\angle A}{360^{\circ}} \times \pi \times(3.5)^{2}+\frac{(\angle B+\angle C)}{360^{\circ}} \times \pi \times(3.5)^{2}\)
\(=\frac{(\angle A+\angle B+\angle C)}{360^{\circ}} \times \pi \times(3.5)^{2} \)
\(=\frac{22}{7} \times \frac{180^{\circ}}{360^{\circ}} \times 3.5 \times 3.5\)
\(\left(\because\right. sum of interior angles of a triangle is 180^{\circ} )\)
\(=\frac{77}{4}=19.25 \mathrm{~m}^{2}\)
(v) (b): Area of the field that cannot be grazed
= Area of MBC - Area of region grazed by all the three animals
= 84 - 19.25 = 64.75 m²

Let r and R be the radii of each smaller circle and larger circle respectively.
We have, \(d=\frac{1}{4} D\)
\(\Rightarrow r=\frac{1}{4} R \Rightarrow r=\frac{1}{4} \times 16 \Rightarrow r=4 \mathrm{~cm}\)
(i) (c): Area of smaller circle = \(\pi r^{2}\)
\(=\frac{22}{7} \dot{\times} 4 \times 4=50.28 \mathrm{~cm}^{2}\)
(ii) (a): Area oflarger circle = \(\pi R^{2}\)
\(=\frac{22}{7} \times 16 \times 16=\frac{5632}{7}=804.57 \mathrm{~cm}^{2}\)
(iii) (b): Area of the black colour region = Area of larger circle - Area of 4 smaller circles
= 804.57 - 4 x 50.28 = 603.45 cm²
(iv) (d): Area of quadrant of a smaller circle
\(=\frac{1}{4} \times 50.28=12.57 \mathrm{~cm}^{2}\)
(v) (c): Area between two concentric circles
\(=\pi\left(R^{2}-r^{2}\right)=\frac{22}{7}\left(5^{2}-2^{2}\right) \)
\(=\frac{22}{7}(25-4)=\frac{22}{7} \times 21=66 \mathrm{~cm}^{2}\)

(i) (b): \(\angle X A C=45^{\circ}\) 
\(\therefore \quad \angle A C D=45^{\circ}\) [Alternate interior angles]
(ii) (b)
(iii) (c) : \(\text { In } \Delta A C D\)
\(\frac{A D}{D C}=\tan 45^{\circ} \)
\(\Rightarrow \frac{100}{D C}=1 \Rightarrow D C=100 \mathrm{~m}\)
(iv) (d): \(\text { In } \Delta A B D, \frac{A D}{B D}=\tan 30^{\circ}\)
\(\Rightarrow \quad \frac{100}{B D}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow \quad B D=100 \sqrt{3} \mathrm{~m}\)
(v) (a): \(\text { In } \Delta A D C\)
\(\frac{A D}{A C}=\sin 45^{\circ} \Rightarrow \frac{100}{A C}=\frac{1}{\sqrt{2}} \Rightarrow A C=100 \sqrt{2} \mathrm{~m}\)

(i) (b): Total height of pole = 8 m
\(\therefore\) BD = AD - AB = (8 - 2)m = 6 m
(ii) (a): \(\text { In } \Delta B D C, \frac{B D}{B C}=\sin 60^{\circ}\)
\(\Rightarrow \quad \frac{6}{B C}=\frac{\sqrt{3}}{2} \)
\(\Rightarrow \quad B C=\frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=4 \sqrt{3} \mathrm{~m}\)
(iii) (d): \(\text { In } \triangle B D C\)
\(\frac{B D}{C D}=\tan 60^{\circ} \Rightarrow \frac{6}{C D}=\sqrt{3} \Rightarrow C D=\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=2 \sqrt{3} \mathrm{~m}\)
(iv) (b) : \(\text { If } \Delta B C D\)
\(\frac{B D}{C D}=\tan \theta \Rightarrow 1=\tan \theta \quad[\because B D=C D] \)
\(\Rightarrow \quad \theta=45^{\circ}\)
(v) (c) : \(\operatorname{In} \Delta B D C, \angle B+\angle D+\angle C=180^{\circ}\)
\(\therefore \quad \angle B=180^{\circ}-60^{\circ}-90^{\circ}=30^{\circ}\)

(i) (c) : Let AC be the length of the ladder.

\(\text { In } \Delta A B C, \frac{B C}{A C}=\sin 30^{\circ} \)
\(\Rightarrow \frac{6}{A C}=\frac{1}{2} \Rightarrow A C=12 \mathrm{~m}\)
(ii) (b): \(\text { In } \Delta A B C, \frac{A B}{A C}=\cos 30^{\circ}\)
\(\Rightarrow \frac{5}{A C}=\frac{\sqrt{3}}{2} \Rightarrow A C=\frac{10}{\sqrt{3}} \mathrm{~m}\)

(iii) (a) : Let BC be the height of window from ground.

\(\text { In } \Delta A B C, \frac{B C}{A B}=\tan 60^{\circ} \)
\(\Rightarrow \frac{B C}{2.5}=\sqrt{3} \)
\(\Rightarrow B C=2.5 \times 1.73=4.325 \mathrm{~m}\)
(iv) (d): Let AB be the horizontal distance between the foot of ladder and wall.

\(\text { In } \Delta A B C, \frac{B C}{A B}=\tan 45^{\circ} \)
\(\Rightarrow \quad \frac{8}{A B}=1 \Rightarrow A B=8 \mathrm{~m}\)
(v) (b): Let the required distance be x.
\(\text { In } \Delta A B C,(9)^{2}=x^{2}+(6)^{2}\)
[By Pythagoras theorem]

\(\Rightarrow 81-36=x^{2} \Rightarrow 45=x^{2} \)
\(\Rightarrow \quad x=3 \sqrt{5} \mathrm{~m}\)

(i) (d): Let h be the height of the pole.
In \(\Delta\)ABC, 

\( \frac{h}{15}=\sin 45^{\circ} \Rightarrow \frac{h}{15}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \quad h =\frac{15}{\sqrt{2}} \mathrm{~m}\)
(ii) (a): Let x be the required distance.
In \(\Delta\)ABC,

\(\frac{x}{15}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \quad x=\frac{15}{\sqrt{2}} \mathrm{~m}\)
(iii) (c) : Let h be the height of the pole.
In right triangle ABC,

\(\frac{h}{15}=\sin 30^{\circ}=\frac{1}{2}\)
\(\Rightarrow \quad h=\frac{15}{2}=7.5 \mathrm{~m}\)
(iv) (d): If 3 m rope is broken, then the length of the rope is 12 m.

\(\text { In } \Delta A B C, \frac{h}{12}=\sin 30^{\circ}=\frac{1}{2} \)
\(\Rightarrow \quad h=\frac{12}{2}=6 \mathrm{~m}\)
(v) (c)

(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)

(i) (c) : Since AE || FD
\(\therefore\) \(\angle\)EAD = \(\angle\)ADF = 30° [Alternate interior angles]

(ii) (b): Since, AE || BC
\(\therefore\) \(\angle\)EAC = \(\angle\)ACB = 60° [Alternate interior angles]
(iii) (a) : In \(\Delta\)ABC,
\(\begin{array}{l} \tan 60^{\circ}=\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{50}{B C} \\ \Rightarrow \quad B C=\frac{50}{\sqrt{3}}=28.90 \mathrm{~m} \end{array}\)
(iv) (c): In \(\Delta\)ADF, \(\tan 30^{\circ}=\frac{A F}{F D}\)
\(\Rightarrow \frac{1}{\sqrt{3}}=\frac{A B-B F}{F D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{50-C D}{\frac{50}{\sqrt{3}}}\)
\(\left[\because F D=B C=\frac{50}{\sqrt{3}}\right] \)
\(\Rightarrow \frac{50}{3}=50-C D \Rightarrow C D=50-\frac{50}{3}=\frac{100}{3}=33.33 \mathrm{~m}\)
(v) (d)

We have, AB = BC = 6\(\sqrt{2}\) m and AC=12m .
(i) (d):\(\because\)  Dis mid point of AC.
\(\therefore\) AD=DC=6m
Now, AB² = BD² + AD² (\(\therefore\) \(\Delta\)ABD is a right triangle)
\(\Rightarrow B D^{2}=(6 \sqrt{2})^{2}-6^{2}=72-36=36 \)
\(\Rightarrow B D=6 \mathrm{~m}\)
(ii) (c) : \(\operatorname{In} \Delta A B D, \sin A=\frac{B D}{A B}=\frac{6}{6 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin A=\sin 45^{\circ} \Rightarrow \angle A=45^{\circ}\)
(iii) (c) : \(\operatorname{In} \Delta B D C, \tan C=\frac{B D}{D C}=\frac{6}{6}\)
\(\Rightarrow \tan C=1=\tan 45^{\circ} \Rightarrow \angle C=45^{\circ}\)
(iv) (d) : \(\sin A=\frac{1}{\sqrt{2}}, \cos C=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)
\(\therefore \quad \sin A+\cos C=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
(v) (c): \((v) \quad(c): \tan C=1, \tan A=\tan 45^{\circ}=1\)
\(\Rightarrow \tan ^{2} C+\tan ^{2} A=1+1=2\)
 

We have, KL = 4 cm, ML = 4\(\sqrt{3}\)m, KM = 8 cm
(i) (a): \(\tan M=\frac{K L}{L M}=\frac{4}{4 \sqrt{3}}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow \tan M=\tan 30^{\circ} \Rightarrow \angle M=30^{\circ}\)
(ii) (c) : \(\tan K=\frac{M L}{K L}=\frac{4 \sqrt{3}}{4}=\sqrt{3}=\tan 60^{\circ}\)
\(\Rightarrow \angle K=60^{\circ}\)
(iii) (b)
(iv) (c)
(v) (a) : \(\frac{\tan ^{2} 45^{\circ}-1}{\tan ^{2} 45^{\circ}+1}=\frac{(1)^{2}-1}{1^{2}+1}=\frac{0}{2}=0\)

\(\because \Delta\)PQR is a right angled triangle.
\(\therefore\) PR² + RQ² = PQ²
\(\Rightarrow P R^{2}=(13)^{2}-(12)^{2}=25 \Rightarrow P R=5 \mathrm{~cm}\)
(i) (c) : \(\cos \theta=\frac{Q R}{P Q}=\frac{12}{13} \)
(ii) (c) : \(\sec \theta=\frac{1}{\cos \theta}=\frac{13}{12} \)
(iii) (c) : \(\tan \theta=\frac{P R}{R Q}=\frac{5}{12}\)
\(\therefore \frac{\tan \theta}{1+\tan ^{2} \theta}=\frac{\frac{5}{12}}{1+\frac{25}{144}}=\frac{\frac{5}{12}}{\frac{169}{144}}=\frac{60}{169}\)
(iv) (a): \(\cot \theta=\frac{1}{\tan \theta}=\frac{12}{5}\)    [Using (1)]
\(\operatorname{cosec} \theta=\frac{P Q}{P R}=\frac{13}{5} \)
\(\therefore \quad \cot ^{2} \theta-\operatorname{cosec}^{2} \theta=\frac{144}{25}-\frac{169}{25}=-1\)
(v) (b): \(\sin ^{2} \theta+\cos ^{2} \theta=1\) (Using identity)

(i) (d) : \(\text { In } \Delta A P Q, \tan \theta=\frac{A Q}{P Q}=\frac{1.2}{1.6}=\frac{3}{4}\)
(ii) (d) : \(\text { In } \Delta P B Q, \cot B=\frac{Q B}{P Q}=\frac{3}{1.6}=\frac{15}{8}\) ...(i)
(iii) (c): \(\text { In } \Delta A P Q, \tan A=\frac{P Q}{A Q}=\frac{1.6}{1.2}=\frac{4}{3}\) ...(ii)
(iv) (d): We have, tan²A + 1 = sec²A
\(\Rightarrow \sec A =\sqrt{\left(\frac{4}{3}\right)^{2}+1} \)
\(=\sqrt{\frac{16}{9}+1}=\sqrt{\frac{25}{9}}=\frac{5}{3}\)
(v) (a): Since \(\operatorname{cosec} B=\sqrt{\cot ^{2} B+1}\)
\(\begin{array}{l} =\sqrt{\left(\frac{15}{8}\right)^{2}+1} \\ =\frac{17}{8} \end{array}\)

(i) (a): We have, AB = 9 m, BC = 3\(\sqrt{3}\) m
In  \(\Delta\)ABC, we have
\(\tan A=\frac{B C}{A B}=\frac{3 \sqrt{3}}{9}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow \tan A=\tan 30^{\circ} \Rightarrow \angle A=30^{\circ}\)
(ii) (c): Similarly, \(\tan C=\frac{A B}{B C}=\frac{9}{3 \sqrt{3}}=\sqrt{3}\)
\(\Rightarrow \tan C=\tan 60^{\circ} \Rightarrow \angle C=60^{\circ}\)
(iii) (d): Since \(\sin A=\frac{B C}{A C} \Rightarrow \sin 30^{\circ}=\frac{B C}{A C}\)
\(\Rightarrow \frac{1}{2}=\frac{3 \sqrt{3}}{A C} \Rightarrow A C=6 \sqrt{3} \mathrm{~m}\)
(iv) (b) : \(\because \angle A=30^{\circ}\) [From (1)]
\(\therefore \quad \cos 2 A=\cos \left(2 \times 30^{\circ}\right)=\cos 60^{\circ}=\frac{1}{2}\) 
(v) (b): \(\because \angle C=60^{\circ}\) [Using (2)]
\(\therefore \quad \sin \left(\frac{C}{2}\right)=\sin \left(\frac{60^{\circ}}{2}\right)=\sin 30^{\circ}=\frac{1}{2}\)