Wait, why is $\angle BDC$ exactly $90^\circ$?

North and West are perpendicular directions on a compass. Any line drawn North and any line drawn West will always meet at a right angle, forming the foundation of our coordinate system.

How did you know $\Delta ADC$ was isosceles?

After calculating $\angle DAC = 25^\circ$, we noticed it matched $\angle ACD$. A triangle with two equal angles must have two equal sides opposite those angles. This is a powerful geometric property that simplifies the problem significantly.

Is there a way to solve this without the isosceles property?

Yes, you could use the Sine Rule in $\Delta ADC$, but recognizing the isosceles property is a much faster, more elegant shortcut that saves time and reduces the chance of calculation errors.

JEE Advanced 1983

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: Four ships $A, B, C$ and $D$ are at sea in the following relative positions : $B$ is on the straight line segment $A C$ , $B$ is due North of $D$ and $D$ is due west of $C$ . The distance between $B$ and $D$ is $2$ km. $∠ B D A = 4 0^{\circ}, ∠ B C D = 2 5^{\circ}$ . What is the distance between $A$ and $D$ ? [Take $sin 2 5^{\circ} = 0.423$ ]

Visualized Solution

Visualizing the Ships' Positions

Ship $B$ is due North of $D$ , so $B D$ is vertical.
Ship $D$ is due West of $C$ , so $D C$ is horizontal.
This forms a right angle: $∠ B D C = 9 0^{\circ}$ .

Positioning Ship $A$

Ship $B$ lies on the straight line segment $A C$ .
We are given $∠ B D A = 4 0^{\circ}$ .
The total angle $∠ A D C = ∠ B D A + ∠ B D C = 4 0^{\circ} + 9 0^{\circ} = 13 0^{\circ}$ .

Calculating $∠ D A C$

Consider the large triangle $Δ A D C$ .
The sum of angles in a triangle is $18 0^{\circ}$ .
$∠ D A C = 18 0^{\circ} - (∠ A D C + ∠ A C D)$ .
$∠ D A C = 18 0^{\circ} - (13 0^{\circ} + 2 5^{\circ}) = 2 5^{\circ}$ .

Identifying the Isosceles Triangle

Notice that $∠ D A C = 2 5^{\circ}$ and $∠ A C D = 2 5^{\circ}$ .
Since two base angles are equal, $Δ A D C$ is an isosceles triangle.
Therefore, the sides opposite to these angles are equal: $A D = C D$ .

Trigonometry in $Δ B D C$

In the right-angled triangle $Δ B D C$ , we know $B D = 2 km$ .
Using the tangent ratio: $tan 2 5^{\circ} = \frac{B D}{C D}$ .
Rearranging gives: $C D = \frac{B D}{t a n 2 5 ^{\circ}} = 2 cot 2 5^{\circ}$ .

Expressing $A D$ in terms of given values

From our isosceles property, $A D = C D$ .
Therefore, $A D = 2 cot 2 5^{\circ}$ .
We are given $sin 2 5^{\circ} = 0.423$ . We need to find $cot 2 5^{\circ}$ .

Converting Sine to Cotangent

Recall the identity: $cot θ = \frac{c o s θ}{s i n θ} = \frac{1 - s i n ^{2} θ}{s i n θ}$ .
Alternatively, $cot θ = \frac{1}{s i n ^{2} θ} - 1$ .
Substitute $sin 2 5^{\circ} = 0.423$ : $A D = 2 \frac{1}{( 0.423 ) ^{2}} - 1$ .

Calculating the Square

First, calculate $(0.423)^{2}$ .
$0.423 \times 0.423 \approx 0.1789$ .
So, the expression becomes $A D = 2 \frac{1}{0.1789} - 1$ .

Evaluating the Square Root Term

Next, compute $\frac{1}{0.1789} \approx 5.589$ .
Subtract $1$ : $5.589 - 1 = 4.589$ .
Now we need the square root: $4.589 \approx 2.142$ .

Final Distance Calculation

Finally, multiply by $2$ : $A D = 2 \times 2.142$ .
$A D = 4.284 km$ .
Rounding to two decimal places, the distance between ship $A$ and $D$ is $4.28 km$ .

00:00 / 00:00

The Sigma Insight: Heights and Distances

Solution Diagram

Analyzing the Setup

Ship

B

is due North of ship

D

, and ship

D

is due West of ship

C

. Since North and West are perpendicular, the line segments

B D

and

D C

meet at a right angle.

This establishes that

∠ B D C = 9 0^{\circ}

, which serves as the fundamental geometric constraint for our calculations.

The Hidden Symmetry

We are given that ships

A

B

, and

C

are collinear. The total angle at

D

, denoted as

∠ A D C

, is the sum of

∠ B D A

and

∠ B D C

Given

∠ B D A = 4 0^{\circ}

and

∠ B D C = 9 0^{\circ}

, we find:

∠ A D C = 4 0^{\circ} + 9 0^{\circ} = 13 0^{\circ}

Now, consider the triangle

Δ A D C

. The sum of angles in any triangle is

18 0^{\circ}

, and we are given

∠ A C D = 2 5^{\circ}

. We calculate the third angle,

∠ D A C

, as follows:

∠ D A C = 18 0^{\circ} - (13 0^{\circ} + 2 5^{\circ}) = 2 5^{\circ}

Because

∠ D A C = ∠ A C D = 2 5^{\circ}

, the triangle

Δ A D C

is an isosceles triangle. Consequently, the sides opposite these angles are equal, meaning

A D = C D

The Trigonometric Bridge

We focus on the right-angled triangle

Δ B D C

. We know the side

B D = 2 km

and the angle

∠ B C D = 2 5^{\circ}

Using the tangent ratio, we have:

tan 2 5^{\circ} = \frac{B D}{C D}

Rearranging for

C D

, we obtain:

C D = \frac{B D}{tan 2 5 ^{\circ}} = 2 cot 2 5^{\circ}

Since

A D = C D

, it follows that

A D = 2 cot 2 5^{\circ}

Final Calculation

Given

sin 2 5^{\circ} = 0.423

, we use the identity

cot θ = \frac{1 - s i n ^{2} θ}{s i n θ}

to find the value of

cot 2 5^{\circ}

A D = 2 \frac{1}{( 0.423 ) ^{2}} - 1

Calculating the values:

(0.423)^{2} \approx 0.1789

\frac{1}{0.1789} \approx 5.589

5.589 - 1 = 4.589 \approx 2.142

Finally, we compute the distance:

A D = 2 \times 2.142 = 4.284 km

Rounding to two decimal places, the distance between ship

A

and ship

D

is $4.28 km$ .

Visualized Solution

Visualizing the Ships' Positions

Positioning Ship $A$

Calculating $∠ D A C$

Identifying the Isosceles Triangle

Trigonometry in $Δ B D C$

Expressing $A D$ in terms of given values

Converting Sine to Cotangent

Calculating the Square

Evaluating the Square Root Term

Final Distance Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Hidden Symmetry

The Trigonometric Bridge

Final Calculation

Similar Questions

From the top of a light-house 60 metres high with its base at the sea-level, the angle of depression of a boat is $1 5^{\circ}$ . The distance of the boat from the foot of the light house is

The angle of elevation of the top of a tower from a point $A$ due north of it is $α$ and from a point $B$ at a distance of $9$ units due west of $A$ is $cos^{- 1} (\frac{3}{13})$ . If the distance of the point $B$ from the tower is $15$ units, then $cot α$ is equal to :

A man from the top of a 100 metres high tower sees a car moving towards the tower at an angle of depression of $3 0^{\circ}$ . After some time, the angle of depression becomes $6 0^{\circ}$ . The distance (in metres) travelled by the car during this time is

If the angle of elevation of a cloud from a point $P$ which is 25 m above a lake be $3 0^{\circ}$ and the angle of depression of reflection of the cloud in the lake from $P$ be $6 0^{\circ}$ , then the height of the cloud (in meters) from the surface of the lake is :

Visualized Solution

Visualizing the Ships' Positions

Positioning Ship .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∠DAC

Identifying the Isosceles Triangle

Trigonometry in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔBDC

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AD in terms of given values

Converting Sine to Cotangent

Calculating the Square

Evaluating the Square Root Term

Final Distance Calculation

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Hidden Symmetry

The Trigonometric Bridge

Final Calculation

Similar Questions

Positioning Ship $A$

Calculating $∠ D A C$

Trigonometry in $Δ B D C$

Expressing $A D$ in terms of given values