**Ans . **

\(60^{\circ}\)

**Explanation :**Let AB be the pole and AC be its shadow.

Let angle of elevation,\(\angle\)ACB=\(\theta\)

Then, AB = 2\(\sqrt{3}\)m AC = 2 m.

Tan \(\theta\)=AB/AC = \(\sqrt[2]{\frac{3}{2}}\)=\(\sqrt{3}\)=>\(\theta\)

So, the angle of elevation is \(60^{\circ}\)

**Ans . **

9.5

**Explanation :**Let AB be the wall and BC be the ladder.

Then, \(\angle\)ACB= \(60^{\circ}\) and BC = 19 m.

Let AC = x metres

AC/BC = cos \(60^{\circ}\) => x/19 = 1/2=>x=19/2 = 9.5

Distance of the foot of the ladder from the wall = 9.5

**Ans . **

20.76 m

**Explanation :**Let AB be the tower and C and D be the points of observation. Then,

AB/AD = tan \(60^{\circ}\) = \(\sqrt{3}\) => AD = AB/\(\sqrt{3}\)= h/\(\sqrt{3}\)

AB/AC = tan \(30^{\circ}\) = 1/\(\sqrt{3}\) AC=AB x\(\sqrt{3}\) = h\(\sqrt{3}\)

CD = (AC-AD) = (h\(\sqrt{3}\)-h/\(\sqrt{3}\))

(h\(\sqrt{3}\)-h/\(\sqrt{3}\)) = 24 => h=12\(\sqrt{3}\)=(12*1.73)=20.76

Hence, the height of the tower is 20.76 m.

**Ans . **

18 m

**Explanation :**Let AB be the tree and AC be the river. Let C and D be the two positions of the man. Then,

\(\angle\)ACB=\(60^{\circ}\),\(\angle\)ADB=\(30^{\circ}\) and CD=36 m.

Let AB=h metres and AC=x metres.

Then, AD=(36+x)metres......(1)

AB/AD=tan \(30^{\circ}\)=1/\(\sqrt{3}\) => h/(36+x)=1/\(\sqrt{3}\) =>h=(36+x)/\(\sqrt{3}\)

AB/AC=tan \(60^{\circ}\)=\(\sqrt{3}\) => h/x=\(\sqrt{3}\)

h=\(\sqrt{3}\)/x .....(2)

From (i) and (ii), we get:

(36+x)/\(\sqrt{3}\)x => x=18 m.

So, the breadth of the river = 18 m.

**Ans . **

5 minutes

**Explanation :**Let AB be the tower and C and D be the two positions of the boat.

Let AB=h, CD=x and AD=y.

h/y=tan \(60^{\circ}\) =\(\sqrt{3}\) => y=h/\(\sqrt{3}\)

h/(x+y)=tan \(30^{\circ}\)=1/\(\sqrt{3}\) => x+y=\(\sqrt{3}\)h

x=(x+y)-y = (\(\sqrt{3}\) h-h/\(\sqrt{3}\) )=2h/\(\sqrt{3}\)

Now, 2h/\(\sqrt{3}\) is covered in 10 min.

h/\(\sqrt{3}\) will be covered in (10*(\(\sqrt{3}\)/2h)*(h/\(\sqrt{3}\)))=5 min

Hence, required time = 5 minutes.

**Ans . **

18 m

**Explanation :**Let AB and CD be the two temples and AC be the river.

Then, AB = 54 m. Let AC = x metres and CD=h metres.

\(\angle\)ACB=\(60^{\circ}\), \(\angle\)EDB=\(30^{\circ}\)

AB/AC=tan \(60^{\circ}\)=\(\sqrt{3}\)

AC=AB/\(\sqrt{3}\)=54/\(\sqrt{3}\)=(54/\(\sqrt{3}\)*\(\sqrt{3}\)*\(\sqrt{3}\))=18m

DE=AC=18\(\sqrt{3}\)

BE/DE=tan \(30^{\circ}\)=1/\(\sqrt{3}\)

BE=(18\(\sqrt{3}\)*1/\(\sqrt{3}\)=18 m

CD=AE=AB-BE=(54-18) m = 36 m.

So, Width of the river = AC = 18\(\sqrt{3}\)m=18*1.73 m=31.14m

Height of the other temple = CD= 18 m