Let AB = h km be the height of the mountain. Let C be a point at a distance of x km from the base of the mountain such that \(\angle\)ACB = 30\(°\) and let D be a point at a distance of 10 km from C along the same line. Then \(\angle\)ADB = 15\(°\) and AD = AC + DC = (x + 10) km

\(In\quad \Delta BAC,\quad tan\quad 30°=\frac { P }{ B } =\frac { AB }{ AC } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } } =\frac { h }{ x } \left[ \because tan 30°=\frac { 1 }{ \sqrt { 3 } } \right] \)
\(\Rightarrow x= h\sqrt { 3 } \)  ......(i)
\(In \Delta BAD, tan 15°=\frac { AB }{ AD } \)
\(\Rightarrow 0.27=\frac { h }{ x+10 } \left[ given, tan 15°=0.27 \right] \)
\(\Rightarrow\)  0.27(x + 10) = h  ......(ii)
On putting x = \(\sqrt { 3 } h\) from Eq. (i) in Eq. (ii), we get
\(0.27(\sqrt { 3 } h+10)=h\)
\(\Rightarrow 0.27\times \sqrt { 3 } h+0.27\times 10=h\)
\(\Rightarrow h(1-0.27\times \sqrt { 3 } )=0.27\times 10\)
\(\Rightarrow h(1-0.27\times 1.732)=2.7\)   \(\left[ \because \sqrt { 3 } =1.732 \right] \)
\(\Rightarrow\) h(1-0.47) = 2.7
\(\Rightarrow 0.53h=2.7\Rightarrow h=\frac { 2.7 }{ 0.53 } =5.09\approx 5km\)
Hence, the height of mountain is 5 km.

34.64 m; 20 m

200 m/min

10.41 m

AD is the lighthouse, Find AD=?
In right ΔABD,
h=x
\(\frac { h }{ x } \)=tan45^o
\(\frac { h }{ 100-x } \)=tan30^o ......(ii)
Solve for h and x.
⇒ \(\frac { h }{ 100-x } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \sqrt { 3 } h\)=100-x
⇒ \(\sqrt { 3 } \)x=100-x   [Using eq (i)]
⇒ (\(\sqrt { 3 } \)+1)x=100 ⇒ x=\(\frac { 100 }{ \sqrt { 3 } +1 } \)
⇒ x=\(\frac { 100(\sqrt { 3 } -1) }{ (\sqrt { 3 } +1)(\sqrt { 3 } -1) } \)
⇒ x=\(\frac { 100(\sqrt { 3 } -1) }{ 2 } =50(\sqrt { 3 } -1)\)m
∴ h=height of lighthouse=\(50(\sqrt { 3 } -1)\)m

Let AB is Ist pole and CD is IInd pole.

CD = 24 m and BD = 15 m
 AE is horizontal line.
In rectangle
 ABDE,
 AE = BD = 15 m
Let CE = x m.
In right \(\Delta \) CEA,
\(\frac { CE }{ AE } =tan{ 30 }^{ o }\Rightarrow \frac { x }{ 15 } =\frac { 1 }{ \sqrt { 3 } } \)m

Let AP and DP be the position of the ladder whose length is 6 m.
In rt.ΔABP, \(\frac { BP }{ AP } =cos60^{ 0 }\Rightarrow \frac { BP }{ 6 } =\frac { 1 }{ 2 } \)
⇒ BP=1/2 x 6=3m
In rt, ΔDCP, \(\frac { PC }{ DP } =cos45^{ 0 }=\frac { PC }{ 6 } =\frac { 1 }{ \sqrt { 2 } } \)
⇒ PC=\(\frac { 6 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 6\sqrt { 2 } }{ 2 } =3\sqrt { 2 } m\)
Distance between two walls
=BP+PC=3+3\(\sqrt { 2 } \)
=3+3 x 1.41
=3(1+1.41)=7.23 m

Given: AB be the tower and AC the flag pole on top of the tower.
ㄥCEB=60^o, ㄥAEB=30^o
To find Height of the tower and the height of the flag pole. Let height of the tower and flag pole be h m and h'm respectively.
Solution: In right ∆ABE
=\(\frac { AB }{ BE } =tan{ 30 }^{ 0 }\Rightarrow \frac { h }{ 9 } =\frac { 1 }{ \sqrt { 3 } } \)
h=\(\frac { 9 }{ \sqrt { 3 } } m=\frac { 9 }{ \sqrt { 3 } } =\frac { 9\sqrt { 3 } }{ 3 } m=3\sqrt { 3 } =5.196\)m  .....(i)
In right ΔCBE,
\(\frac { CB }{ BE } =tan600\Rightarrow \frac { h+{ h }^{ ' } }{ BE } =tan60^{ 0 }\)
\(\frac { h+h' }{ BE } =\sqrt { 3 } \)
⇒ h+h'=\(9\sqrt { 3 } \)               .......(ii)
\(h'=\frac { 27-9 }{ \sqrt { 3 } } =\frac { 18\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } =\frac { 18\sqrt { 3 } }{ 3 } =6\sqrt { 3 } \)m
=6 x 1.732 m=10.392 m
Height of flag pole mounted on tower=10.392 m.

Let AB is statue, BC is pedestal and BC=x m, CD=y m.
In right ΔBCD, ΔBCD, \(\frac { BC }{ CD } \)=tan 45^o
⇒, \(\frac { x }{ y } \)=1 ⇒ x=y     .....(i)
In right ΔACD, \(\frac { AC }{ CD } \)=tan 600
⇒ \(\frac { x+1.46 }{ y } =\sqrt { 3 } \)
⇒ \(\frac { x+1.46 }{ x } \)=1.73    [Using (i)]
⇒ x+1.46=1.73x ⇒ 0.73x=1.46 ⇒ x=2

Let AB be the tower and AC be the flag staff on the tower. Let D be a point on the ground such that the angles of elevation of foot A and top C of the flag staff are 45\(°\) and 60\(°\) respectively.
Then, we have AB = 20 m, \(\angle ADB=45° and \angle CDB=60°\)

In right angled \(\Delta ABD,\)
\(tan\quad 45°=\frac { AB }{ BD } \left[ \because tan\theta =\frac { P }{ B } \right] \)
\(\Rightarrow 1=\frac { 20 }{ BD } \)
\(\Rightarrow BD=20\quad m \left[ \because tan45°=1 \right] \)
and in right angled \(\Delta CBD,\)
\(tan 60°=\frac { BC }{ BD } \)
\(\sqrt { 3 } =\frac { BC }{ 20 } \left[ \because tan60°=\sqrt { 3 } \right] \)
\(\Rightarrow BC=20\sqrt { 3 } =20\times 1.732 \left[ \because \sqrt { 3 } =1,732 \right] \)
= 34.64 m (approx)
Now, AC = BC - AB = 34.64 - 20
=14.64 m (approx)
Hence, the height of the fleg staff is 14.64 m.

Let AB be the tree whose part AC breaks and touches the ground at D
Then,  BD = \(2\sqrt { 3 } m\)
and AC = CD
In right angled \(\Delta \)CBD,
\(cos 60°=\frac { B }{ H } =\frac { BD }{ CD } \)
\(\Rightarrow \frac { 1 }{ 2 } =\frac { 2\sqrt { 3 } }{ CD } \)
\(\left[ \because cos 60°=\frac { 1 }{ 2 } and BD=2\sqrt { 3 } m \right] \)

\(\Rightarrow\) \(CD=2\times 2\sqrt { 3 } =4\sqrt { 3 } \)
\(=4\times 1.732=6.928\ m [\because \sqrt { 3 } =1.732]\)
\(\therefore\)  AC = CD = 6.928 m
Again, in right angled \(\Delta \)CBD,
\(tan\quad 60°=\frac { P }{ B } =\frac { BC }{ BD } \)
\(\Rightarrow \sqrt { 3 } =\frac { BC }{ 2\sqrt { 3 } } [\because tan60°=\sqrt { 3 } and BD=2\sqrt { 3 } m]\)
\(\Rightarrow BC=\sqrt { 3 } \times 2\sqrt { 3 } =6 m\)
Now, AB = AC + BC
= 6.928 + 6 = 12.928 m (approx)
Hence, the height of the tree is 12.928 m.

Let BC be the width of the river and A, B be the ends of river such that
AB = 150 m = Length of the bridge

and  \(\angle\)BAC = 45 \(°\)
In right angled \(\Delta ACB,\)
\(sin\quad 45°=\frac { P }{ H } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 2 } } =\frac { BC }{ AB } =\frac { BC }{ 150 } \)
\(\therefore BC=\frac { 150 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } \) [by rationalising]
\(=\frac { 150 }{ 2 } \sqrt { 2 } =75\sqrt { 2 } \)
\(\\ =75\times 1.414\quad [\because \sqrt { 2 } =1.414]\)
=106.05 m (approx)
Hence, width of the river is 106.05 m.

Let AB be the tower whose height is 300\(\sqrt { 3 } \) m, Again, let C be the point at a distance of 300 m from the foot of the tower, i.e. AC = 300 m.

Here, the angle of elevation is unknown, so let it be \(\theta \) 
Since, here base and perpendicular are given.
So, in right angled \(\Delta \)BAC,
\(tan \theta =\frac { perpendicular }{ Base } =\frac { AB }{ AC } =\frac { 300\sqrt { 3 } }{ 300 } \)
\(\Rightarrow tan \theta =\sqrt { 3 } =tan\quad 60°\)
\(\\ \therefore \theta =60°\)
Hence, the required angle of elevation is 60\(°\)

(i) Let OA be the tree of height bm.
Given PQ = 100m and angles of elevation are
\(\angle APO={ 30 }^{ \circ } \angle OQA={ 45 }^{ \circ }\)
In right angled  \(\Delta POA,\)  \(\tan { { 30 }^{ \circ } } =\frac { OA }{ OP } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } } =\frac { b }{ OP } \Rightarrow OP=\sqrt { 3b } .....(i)\)

Now, in right angled \(\Delta QOA,\)
\(\tan { { 45 }^{ \circ } } =\frac { OA }{ OQ } \Rightarrow 1=\frac { b }{ OQ } \Rightarrow OQ=b\ ....(ii)\)
On adding Eqs.  (i) and (ii), we get
\(OP+OQ=\sqrt { 3b } +b\)
\(\Rightarrow PQ=(\sqrt { 3 } +1)b\)
\(\Rightarrow 100=(\sqrt { 3 } +1)b[\because PQ=100m, give]\)
\(\\ \Rightarrow b=\frac { 100 }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \)
\( \Rightarrow b=50(\sqrt { 3 } -1)=50(1.732-1) [\because \sqrt { 3 } =1.732]\)
\( \Rightarrow b=36.6m\)
Hence, height of the tree is 36.6m
(ii) 99.99 m,
(iii) Point P

Given: tower AB height h m . BC is the shadow when sun's altitude is 45^o and shadow is BD, when the sun's altitude is 30^o.
To find: The height of tower
Solution: In rt. ΔABC,
\(\frac { AB }{ BC } =tan{ 45 }^{ 0 }\Rightarrow \frac { h }{ y } =1\)
or h = y .....(i)
⇒ \(\frac { AB }{ BD } =\frac { h }{ y+2x } \) = tan30^o
⇒ \(\frac { h }{ y+2x } =\frac { 1 }{ \sqrt { 3 } } \)
\(\sqrt { 3 } \)h = y + 2x ......(ii)
From (i) and (ii), we have
\(\sqrt { 3 } \)h = h +2 x ⇒ \(\sqrt { 3 } \)h - h = 2x
⇒ (\(\sqrt { 3 } \) - 1)h = 2x
⇒ h = \(\frac { 2x }{ (\sqrt { 3 } -1) } =\frac { 2x(\sqrt { 3 } +1) }{ (\sqrt { 3 } -1)(\sqrt { 3 } +1) } \)
= \(\frac { 2x(\sqrt { 3 } +1) }{ (\sqrt { 3 } )^{ 2 }-1^{ 2 } } =\frac { 2x(\sqrt { 3 } +1) }{ 3-1 } \)
= \(\frac { 2x(\sqrt { 3 } +1) }{ 2 } =x(\sqrt { 3 } +1)\) m