EXAMPLES
Some Applications Of Trigonometry • 7 Questions
Question 1
Hint available
A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.
Key Idea
Use the definition of the tangent function in a right‑angled triangle: \(\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}\). Here the opposite side is the height of the tower and the adjacent side is the horizontal distance (15 m).
Step-by-Step Solution
1. Draw a right‑angled triangle:
- The foot of the tower, the point on the ground, and the top of the tower form a right‑angled triangle.
- Let \(h\) be the height of the tower (opposite side).
- The distance from the point to the foot of the tower is given as \(15\) m (adjacent side).
- The angle of elevation at the point is \(\theta = 60^{\circ}\).
2. Apply the definition of tangent:
$$\tan 60^{\circ} = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{15}$$
3. Use the known value \(\tan 60^{\circ} = \sqrt{3}\):
$$\sqrt{3} = \frac{h}{15}$$
4. Solve for \(h\):
$$h = 15 \times \sqrt{3}$$
5. Evaluate numerically (if required):
$$h \approx 15 \times 1.732 = 25.98 \text{ m} \approx 26 \text{ m}$$
Thus the height of the tower is \(15\sqrt{3}\) m (approximately 26 m).
- The foot of the tower, the point on the ground, and the top of the tower form a right‑angled triangle.
- Let \(h\) be the height of the tower (opposite side).
- The distance from the point to the foot of the tower is given as \(15\) m (adjacent side).
- The angle of elevation at the point is \(\theta = 60^{\circ}\).
2. Apply the definition of tangent:
$$\tan 60^{\circ} = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{15}$$
3. Use the known value \(\tan 60^{\circ} = \sqrt{3}\):
$$\sqrt{3} = \frac{h}{15}$$
4. Solve for \(h\):
$$h = 15 \times \sqrt{3}$$
5. Evaluate numerically (if required):
$$h \approx 15 \times 1.732 = 25.98 \text{ m} \approx 26 \text{ m}$$
Thus the height of the tower is \(15\sqrt{3}\) m (approximately 26 m).
Question 2
Hint available
An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3m below the top of the pole to undertake the repair work (see Fig. 9.5). What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (You may take 3 = 1.73)
Key Idea
Use the basic trigonometric ratios for a right‑angled triangle: \(\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}\) and \(\tan\theta = \frac{\text{opposite}}{\text{adjacent}}\). Here the ladder, the ground and the pole form a right‑angled triangle with the ladder as the hypotenuse, the vertical distance to be reached as the opposite side and the horizontal distance from the pole as the adjacent side.
Step-by-Step Solution
1. Determine the vertical distance to be reached.
The pole is 5 m high and the electrician must work 1.3 m below the top, so the required vertical height from the ground is
$$h = 5 - 1.3 = 3.7\ \text{m}.$$
2. Use the given angle (60°) with the horizontal.
The ladder makes an angle \(\theta = 60^{\circ}\) with the ground.
3. Find the length of the ladder (hypotenuse).
Using the sine ratio:
$$\sin 60^{\circ} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{L}.$$
Since \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\) and \(\sqrt{3} \approx 1.73\),
$$\sin 60^{\circ} = \frac{1.73}{2} = 0.865.$$
Hence
$$L = \frac{h}{\sin 60^{\circ}} = \frac{3.7}{0.865} \approx 4.28\ \text{m}.$$
(Exact form: \(L = \frac{3.7 \times 2}{\sqrt{3}} = \frac{7.4}{\sqrt{3}} = \frac{7.4\sqrt{3}}{3}\).)
4. Find the horizontal distance from the foot of the ladder to the pole.
Using the cosine ratio or tangent ratio:
- Cosine method:
$$\cos 60^{\circ} = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{d}{L} \;\Rightarrow\; d = L\cos 60^{\circ} = L \times \frac{1}{2} = \frac{L}{2}.$$
Substituting \(L \approx 4.28\) gives
$$d \approx \frac{4.28}{2} = 2.14\ \text{m}.$$
- Tangent method (check):
$$\tan 60^{\circ} = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d} \;\Rightarrow\; d = \frac{h}{\tan 60^{\circ}} = \frac{3.7}{1.73} \approx 2.14\ \text{m}.$$
5. State the final answers.
- Length of ladder required: approximately \(4.28\) m (or \(4.3\) m to one decimal place).
- Distance of the foot of the ladder from the pole: approximately \(2.14\) m (or \(2.1\) m to one decimal place).
The pole is 5 m high and the electrician must work 1.3 m below the top, so the required vertical height from the ground is
$$h = 5 - 1.3 = 3.7\ \text{m}.$$
2. Use the given angle (60°) with the horizontal.
The ladder makes an angle \(\theta = 60^{\circ}\) with the ground.
3. Find the length of the ladder (hypotenuse).
Using the sine ratio:
$$\sin 60^{\circ} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{L}.$$
Since \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\) and \(\sqrt{3} \approx 1.73\),
$$\sin 60^{\circ} = \frac{1.73}{2} = 0.865.$$
Hence
$$L = \frac{h}{\sin 60^{\circ}} = \frac{3.7}{0.865} \approx 4.28\ \text{m}.$$
(Exact form: \(L = \frac{3.7 \times 2}{\sqrt{3}} = \frac{7.4}{\sqrt{3}} = \frac{7.4\sqrt{3}}{3}\).)
4. Find the horizontal distance from the foot of the ladder to the pole.
Using the cosine ratio or tangent ratio:
- Cosine method:
$$\cos 60^{\circ} = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{d}{L} \;\Rightarrow\; d = L\cos 60^{\circ} = L \times \frac{1}{2} = \frac{L}{2}.$$
Substituting \(L \approx 4.28\) gives
$$d \approx \frac{4.28}{2} = 2.14\ \text{m}.$$
- Tangent method (check):
$$\tan 60^{\circ} = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d} \;\Rightarrow\; d = \frac{h}{\tan 60^{\circ}} = \frac{3.7}{1.73} \approx 2.14\ \text{m}.$$
5. State the final answers.
- Length of ladder required: approximately \(4.28\) m (or \(4.3\) m to one decimal place).
- Distance of the foot of the ladder from the pole: approximately \(2.14\) m (or \(2.1\) m to one decimal place).
Question 3
Hint available
An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?
Key Idea
Use the definition of tangent in a right‑angled triangle: \(\tan \theta = \dfrac{\text{opposite side}}{\text{adjacent side}}\). Here the opposite side is the vertical height of the chimney above the observer's eye level, and the adjacent side is the horizontal distance between the observer and the chimney.
Step-by-Step Solution
1. Draw a right‑angled triangle
- Let point \(O\) be the observer's eye (1.5 m above the ground).
- Let point \(C\) be the top of the chimney.
- Let point \(B\) be the foot of the chimney on the ground.
- \(OB\) is the horizontal distance = 28.5 m.
- \(OC\) is the line of sight making an angle of elevation \(45^{\circ}\) with the horizontal.
- \(BC\) is the vertical height of the chimney above the ground (what we need).
2. Apply the tangent definition
\[\tan 45^{\circ} = \frac{\text{height of chimney above observer's eye}}{\text{horizontal distance}}\]
Since \(\tan 45^{\circ}=1\),
\[1 = \frac{BC - 1.5}{28.5}\]
where \(BC\) is the total height of the chimney and \(1.5\) m is the observer's height.
3. Solve for the unknown height
\[BC - 1.5 = 28.5 \times 1 = 28.5\]
\[BC = 28.5 + 1.5 = 30 \text{ m}\]
4. State the answer
The height of the chimney is \(30\) metres.
- Let point \(O\) be the observer's eye (1.5 m above the ground).
- Let point \(C\) be the top of the chimney.
- Let point \(B\) be the foot of the chimney on the ground.
- \(OB\) is the horizontal distance = 28.5 m.
- \(OC\) is the line of sight making an angle of elevation \(45^{\circ}\) with the horizontal.
- \(BC\) is the vertical height of the chimney above the ground (what we need).
2. Apply the tangent definition
\[\tan 45^{\circ} = \frac{\text{height of chimney above observer's eye}}{\text{horizontal distance}}\]
Since \(\tan 45^{\circ}=1\),
\[1 = \frac{BC - 1.5}{28.5}\]
where \(BC\) is the total height of the chimney and \(1.5\) m is the observer's height.
3. Solve for the unknown height
\[BC - 1.5 = 28.5 \times 1 = 28.5\]
\[BC = 28.5 + 1.5 = 30 \text{ m}\]
4. State the answer
The height of the chimney is \(30\) metres.
Question 4
Hint available
From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (You may take 3 = 1.732)
Key Idea
Use the definition of tangent in a right‑angled triangle: \(\tan\theta = \dfrac{\text{opposite side}}{\text{adjacent side}}\). Apply it first to the building (height 10 m) and then to the total height (building + flagstaff). Solve the two equations to obtain the distance and the flagstaff length.
Step-by-Step Solution
1. Let the horizontal distance from point \(P\) to the foot of the building be \(x\) metres.
2. For the top of the building (height = 10 m) the angle of elevation is \(30^{\circ}\).
$$\tan 30^{\circ}=\frac{10}{x}$$
Using \(\tan 30^{\circ}=\frac{1}{\sqrt{3}}\) and \(\sqrt{3}=1.732\),
$$\frac{1}{1.732}=\frac{10}{x}\quad\Rightarrow\quad x=10\times 1.732=10\sqrt{3}\;\text{m}$$
Hence, \(x = 10\sqrt{3}\) m \(\approx 17.32\) m.
3. Let the length of the flagstaff be \(h\) metres. The total height of the flagstaff top is \(10+h\) metres.
4. The angle of elevation to the top of the flagstaff is \(45^{\circ}\).
$$\tan 45^{\circ}=\frac{10+h}{x}$$
Since \(\tan 45^{\circ}=1\),
$$1=\frac{10+h}{x}\quad\Rightarrow\quad 10+h = x$$
5. Substitute the value of \(x\) from step 2:
$$10+h = 10\sqrt{3}\quad\Rightarrow\quad h = 10\sqrt{3}-10 = 10(\sqrt{3}-1)\;\text{m}$$
Numerically, \(h = 10(1.732-1)=10\times0.732=7.32\) m.
6. Therefore,
- Distance of the building from \(P\): \(x = 10\sqrt{3}\) m \(\approx 17.32\) m.
- Length of the flagstaff: \(h = 10(\sqrt{3}-1)\) m \(\approx 7.32\) m.
2. For the top of the building (height = 10 m) the angle of elevation is \(30^{\circ}\).
$$\tan 30^{\circ}=\frac{10}{x}$$
Using \(\tan 30^{\circ}=\frac{1}{\sqrt{3}}\) and \(\sqrt{3}=1.732\),
$$\frac{1}{1.732}=\frac{10}{x}\quad\Rightarrow\quad x=10\times 1.732=10\sqrt{3}\;\text{m}$$
Hence, \(x = 10\sqrt{3}\) m \(\approx 17.32\) m.
3. Let the length of the flagstaff be \(h\) metres. The total height of the flagstaff top is \(10+h\) metres.
4. The angle of elevation to the top of the flagstaff is \(45^{\circ}\).
$$\tan 45^{\circ}=\frac{10+h}{x}$$
Since \(\tan 45^{\circ}=1\),
$$1=\frac{10+h}{x}\quad\Rightarrow\quad 10+h = x$$
5. Substitute the value of \(x\) from step 2:
$$10+h = 10\sqrt{3}\quad\Rightarrow\quad h = 10\sqrt{3}-10 = 10(\sqrt{3}-1)\;\text{m}$$
Numerically, \(h = 10(1.732-1)=10\times0.732=7.32\) m.
6. Therefore,
- Distance of the building from \(P\): \(x = 10\sqrt{3}\) m \(\approx 17.32\) m.
- Length of the flagstaff: \(h = 10(\sqrt{3}-1)\) m \(\approx 7.32\) m.
Question 5
Hint available
The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.
Key Idea
Use the definition of tangent in a right‑angled triangle: \(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}\). For a tower of height \(h\) and its shadow length \(s\), \(\tan\theta = \dfrac{h}{s}\). Set up two equations for the two given altitudes and use the given difference of shadows to solve for \(h\).
Step-by-Step Solution
1. Let \(h\) be the height of the tower (in metres).\
2. Let \(s_1\) be the length of the shadow when the Sun’s altitude is \(30^{\circ}\) and \(s_2\) be the length when the altitude is \(60^{\circ}\).\
3. From the statement, \(s_1 = s_2 + 40\) (the shadow is 40 m longer at 30°).\
4. Using the definition of tangent:
\[ \tan 30^{\circ} = \frac{h}{s_1} \quad\text{and}\quad \tan 60^{\circ} = \frac{h}{s_2}. \]
5. Substitute the known values of the tangents:
\[ \frac{1}{\sqrt{3}} = \frac{h}{s_1} \quad\Rightarrow\quad h = \frac{s_1}{\sqrt{3}} \]
\[ \sqrt{3} = \frac{h}{s_2} \quad\Rightarrow\quad h = s_2\sqrt{3}. \]
6. Equate the two expressions for \(h\):
\[ \frac{s_1}{\sqrt{3}} = s_2\sqrt{3}. \]
7. Replace \(s_1\) by \(s_2 + 40\):
\[ \frac{s_2 + 40}{\sqrt{3}} = s_2\sqrt{3}. \]
8. Multiply both sides by \(\sqrt{3}\):
\[ s_2 + 40 = 3s_2. \]
9. Solve for \(s_2\):
\[ 3s_2 - s_2 = 40 \Rightarrow 2s_2 = 40 \Rightarrow s_2 = 20\ \text{m}. \]
10. Find the height using \(h = s_2\sqrt{3}\):
\[ h = 20\sqrt{3}\ \text{m}. \]
11. Numerically, \(h \approx 20 \times 1.732 = 34.64\ \text{m}. \]
2. Let \(s_1\) be the length of the shadow when the Sun’s altitude is \(30^{\circ}\) and \(s_2\) be the length when the altitude is \(60^{\circ}\).\
3. From the statement, \(s_1 = s_2 + 40\) (the shadow is 40 m longer at 30°).\
4. Using the definition of tangent:
\[ \tan 30^{\circ} = \frac{h}{s_1} \quad\text{and}\quad \tan 60^{\circ} = \frac{h}{s_2}. \]
5. Substitute the known values of the tangents:
\[ \frac{1}{\sqrt{3}} = \frac{h}{s_1} \quad\Rightarrow\quad h = \frac{s_1}{\sqrt{3}} \]
\[ \sqrt{3} = \frac{h}{s_2} \quad\Rightarrow\quad h = s_2\sqrt{3}. \]
6. Equate the two expressions for \(h\):
\[ \frac{s_1}{\sqrt{3}} = s_2\sqrt{3}. \]
7. Replace \(s_1\) by \(s_2 + 40\):
\[ \frac{s_2 + 40}{\sqrt{3}} = s_2\sqrt{3}. \]
8. Multiply both sides by \(\sqrt{3}\):
\[ s_2 + 40 = 3s_2. \]
9. Solve for \(s_2\):
\[ 3s_2 - s_2 = 40 \Rightarrow 2s_2 = 40 \Rightarrow s_2 = 20\ \text{m}. \]
10. Find the height using \(h = s_2\sqrt{3}\):
\[ h = 20\sqrt{3}\ \text{m}. \]
11. Numerically, \(h \approx 20 \times 1.732 = 34.64\ \text{m}. \]
Question 6
Hint available
The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi- storeyed building and the distance between the two buildings.
Key Idea
Use the relationship between angles of depression and elevation (alternate interior angles) and apply the definition of tangent in right‑angled triangles: \(\tan\theta = \frac{\text{opposite}}{\text{adjacent}}\). Set up two equations for the unknown height \(H\) of the multi‑storeyed building and the horizontal distance \(d\) between the buildings, then solve simultaneously.
Step-by-Step Solution
1. Draw a diagram (not shown). Let:
- \(P\) be the top of the multi‑storeyed building (height \(H\) from ground).
- \(T\) be the top of the 8 m building.
- \(B\) be the bottom (ground) of the 8 m building.
- \(d\) be the horizontal distance between the two buildings.
2. Convert angles of depression to angles of elevation:
- Angle of depression to the top \(T\) = 30° ⇒ angle of elevation from \(T\) to \(P\) = 30°.
- Angle of depression to the bottom \(B\) = 45° ⇒ angle of elevation from \(B\) to \(P\) = 45°.
3. Apply the tangent definition to the two right‑angled triangles:
- For triangle \(P T\) (top to top):
$$\tan 30^{\circ}=\frac{\text{vertical difference}}{\text{horizontal distance}}=\frac{H-8}{d}$$
Since \(\tan30^{\circ}=\frac{1}{\sqrt3}\),
$$\frac{H-8}{d}=\frac{1}{\sqrt3}\quad\Rightarrow\quad H-8=\frac{d}{\sqrt3}\tag{1}$$
- For triangle \(P B\) (top to bottom):
$$\tan 45^{\circ}=\frac{H}{d}$$
Because \(\tan45^{\circ}=1\),
$$\frac{H}{d}=1\quad\Rightarrow\quad H=d\tag{2}$$
4. Substitute \(H=d\) from (2) into (1):
$$d-8=\frac{d}{\sqrt3}\
\Rightarrow\; \sqrt3\,(d-8)=d\
\Rightarrow\; \sqrt3\,d-8\sqrt3=d\
\Rightarrow\; d(\sqrt3-1)=8\sqrt3$$
5. Solve for \(d\):
$$d=\frac{8\sqrt3}{\sqrt3-1}\
=\frac{8\sqrt3(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}\
=\frac{8\sqrt3(\sqrt3+1)}{3-1}\
=4\sqrt3(\sqrt3+1)\
=4(3+\sqrt3)\
=12+4\sqrt3\;\text{metres}$$
Numerically, \(d\approx12+4(1.732)=12+6.928\approx18.93\) m.
6. Find the height \(H\) using (2):
$$H=d=12+4\sqrt3\;\text{metres}\approx18.93\;\text{m}$$
7. Answer: Height of the multi‑storeyed building = \(12+4\sqrt3\) m (≈ 18.93 m). Distance between the two buildings = \(12+4\sqrt3\) m (≈ 18.93 m).
- \(P\) be the top of the multi‑storeyed building (height \(H\) from ground).
- \(T\) be the top of the 8 m building.
- \(B\) be the bottom (ground) of the 8 m building.
- \(d\) be the horizontal distance between the two buildings.
2. Convert angles of depression to angles of elevation:
- Angle of depression to the top \(T\) = 30° ⇒ angle of elevation from \(T\) to \(P\) = 30°.
- Angle of depression to the bottom \(B\) = 45° ⇒ angle of elevation from \(B\) to \(P\) = 45°.
3. Apply the tangent definition to the two right‑angled triangles:
- For triangle \(P T\) (top to top):
$$\tan 30^{\circ}=\frac{\text{vertical difference}}{\text{horizontal distance}}=\frac{H-8}{d}$$
Since \(\tan30^{\circ}=\frac{1}{\sqrt3}\),
$$\frac{H-8}{d}=\frac{1}{\sqrt3}\quad\Rightarrow\quad H-8=\frac{d}{\sqrt3}\tag{1}$$
- For triangle \(P B\) (top to bottom):
$$\tan 45^{\circ}=\frac{H}{d}$$
Because \(\tan45^{\circ}=1\),
$$\frac{H}{d}=1\quad\Rightarrow\quad H=d\tag{2}$$
4. Substitute \(H=d\) from (2) into (1):
$$d-8=\frac{d}{\sqrt3}\
\Rightarrow\; \sqrt3\,(d-8)=d\
\Rightarrow\; \sqrt3\,d-8\sqrt3=d\
\Rightarrow\; d(\sqrt3-1)=8\sqrt3$$
5. Solve for \(d\):
$$d=\frac{8\sqrt3}{\sqrt3-1}\
=\frac{8\sqrt3(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}\
=\frac{8\sqrt3(\sqrt3+1)}{3-1}\
=4\sqrt3(\sqrt3+1)\
=4(3+\sqrt3)\
=12+4\sqrt3\;\text{metres}$$
Numerically, \(d\approx12+4(1.732)=12+6.928\approx18.93\) m.
6. Find the height \(H\) using (2):
$$H=d=12+4\sqrt3\;\text{metres}\approx18.93\;\text{m}$$
7. Answer: Height of the multi‑storeyed building = \(12+4\sqrt3\) m (≈ 18.93 m). Distance between the two buildings = \(12+4\sqrt3\) m (≈ 18.93 m).
Question 7
Hint available
From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.
Key Idea
Use the definition of angle of depression and the tangent ratio in right‑angled triangles. For an angle of depression $\theta$, $\tan\theta = \dfrac{\text{height}}{\text{horizontal distance}}$. Hence horizontal distances to each bank are obtained as $\dfrac{h}{\tan\theta}$ and summed to get the total width.
Step-by-Step Solution
1. Draw a horizontal line through the point $P$ on the bridge (height $h=3\,\text{m}$).\
2. Drop perpendiculars from $P$ to the two banks meeting the river banks at $A$ (near bank) and $B$ (far bank).\
3. $\angle$ of depression to the near bank $A$ is $45^{\circ}$, to the far bank $B$ is $30^{\circ}$.\
4. In right‑angled triangle $\triangle P A Q$ (where $Q$ is the foot of the perpendicular on the near bank),\
$$\tan 45^{\circ}=\frac{\text{height}}{PA}=\frac{3}{PA}\;\Rightarrow\;PA=\frac{3}{\tan45^{\circ}}=\frac{3}{1}=3\,\text{m}.$$\
5. In right‑angled triangle $\triangle P B R$ (where $R$ is the foot of the perpendicular on the far bank),\
$$\tan 30^{\circ}=\frac{\text{height}}{PB}=\frac{3}{PB}\;\Rightarrow\;PB=\frac{3}{\tan30^{\circ}}=\frac{3}{\frac{1}{\sqrt{3}}}=3\sqrt{3}\,\text{m}.$$\
6. The total width of the river $= PA + PB = 3 + 3\sqrt{3}=3(1+\sqrt{3})\,\text{m}.$
2. Drop perpendiculars from $P$ to the two banks meeting the river banks at $A$ (near bank) and $B$ (far bank).\
3. $\angle$ of depression to the near bank $A$ is $45^{\circ}$, to the far bank $B$ is $30^{\circ}$.\
4. In right‑angled triangle $\triangle P A Q$ (where $Q$ is the foot of the perpendicular on the near bank),\
$$\tan 45^{\circ}=\frac{\text{height}}{PA}=\frac{3}{PA}\;\Rightarrow\;PA=\frac{3}{\tan45^{\circ}}=\frac{3}{1}=3\,\text{m}.$$\
5. In right‑angled triangle $\triangle P B R$ (where $R$ is the foot of the perpendicular on the far bank),\
$$\tan 30^{\circ}=\frac{\text{height}}{PB}=\frac{3}{PB}\;\Rightarrow\;PB=\frac{3}{\tan30^{\circ}}=\frac{3}{\frac{1}{\sqrt{3}}}=3\sqrt{3}\,\text{m}.$$\
6. The total width of the river $= PA + PB = 3 + 3\sqrt{3}=3(1+\sqrt{3})\,\text{m}.$