From the previous step, we have AD = (3)/(tan(30^°)). We know that tan(30^°) = (1)/(√3). Substituting this value, we find that AD = 3√3 m.

Now, consider the right-angled triangle PDB. We know the opposite side PD = 3 m and the angle PBD = 45^°. We want to find the adjacent side BD. Using the tangent function again, we have tan(45^°) = (3)/(BD).

Now, consider the right-angled triangle PDB. We know the opposite side PD = 3 m and the angle PBD = 45^°. We want to find the adjacent side BD. Using the tangent function again, we have tan(45^°) = (3)/(BD).

From the equation BD = (3)/(tan(45^°)), and knowing that tan(45^°) = 1, we can calculate BD. Thus, BD = 3 m.

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^\circ$ and $45^\circ$ , respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.

Answer: The width of the river is $3(\sqrt{3} + 1)$ meters.

Step-by-step solution

Step 1: Visualize the Problem and Define Variables

We first visualize the scenario. Let $P$ be the point on the bridge where the observation is made. Let $D$ be the point on the river bank directly below $P$ . Let $A$ and $B$ be points on the opposite banks. The problem states that the height of the bridge from the banks is $3 \text{ m}$ , so $PD = 3 \text{ m}$ . This forms two right-angled triangles, $\triangle PDA$ and $\triangle PDB$ .

Step 2: Identify Angles of Depression and Alternate Interior Angles

The angles of depression are given from point $P$ to the points $A$ and $B$ on the banks. Since the horizontal line through $P$ is parallel to the river banks, the angle of depression to point $A$ is equal to the alternate interior angle $\angle PAD = 30^\circ$ . Similarly, the angle of depression to point $B$ is equal to the alternate interior angle $\angle PBD = 45^\circ$ .

Step 3: Calculate Distance AD Using Trigonometry in $\triangle PDA$

In the right-angled triangle $\triangle PDA$ , we know the opposite side $PD = 3 \text{ m}$ and the angle $\angle PAD = 30^\circ$ . We want to find the adjacent side $AD$ . The tangent function relates these quantities: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ . Substituting the known values, we get $\tan(30^\circ) = \frac{3}{AD}$ .

Step 4: Solve for AD

From the previous step, we have $AD = \frac{3}{\tan(30^\circ)}$ . We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$ . Substituting this value, we find that $AD = 3\sqrt{3} \text{ m}$ .

Step 5: Calculate Distance BD Using Trigonometry in $\triangle PDB$

Now, consider the right-angled triangle $\triangle PDB$ . We know the opposite side $PD = 3 \text{ m}$ and the angle $\angle PBD = 45^\circ$ . We want to find the adjacent side $BD$ . Using the tangent function again, we have $\tan(45^\circ) = \frac{3}{BD}$ .

Step 6: Solve for BD

From the equation $BD = \frac{3}{\tan(45^\circ)}$ , and knowing that $\tan(45^\circ) = 1$ , we can calculate $BD$ . Thus, $BD = 3 \text{ m}$ .

Step 7: Calculate the Width of the River

The width of the river is the sum of the distances $AD$ and $BD$ , since $D$ lies between $A$ and $B$ . Adding the values we calculated for $AD$ and $BD$ , we get the total width $AB = 3\sqrt{3} + 3 \text{ m}$ . This can be factored as $3(\sqrt{3} + 1) \text{ m}$ .

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From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30∘30^\circ30∘ and 45∘45^\circ45∘, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.