# Video: CBSE Class X • Pack 2 • 2017 • Question 27

CBSE Class X • Pack 2 • 2017 • Question 27

04:52

### Video Transcript

An aeroplane is flying at a height of 300 metres above ground. Flying at this height, the angles of depression from the aeroplane to the nearest points on opposite banks of the river are 45 degrees and 60 degrees, respectively. Find the width of the river. Use the square root of three is equal to 1.732.

We can start with a sketch. We have an aeroplane that is flying at a height of 300 metres. We can imagine the flight path of the plane being parallel to the water below — this is what a 45-degree angle of depression would look like — to the nearest point on the bank of the river. We’ll call it 𝐸. This angle also measures 45 degrees because these two angles are alternate interior angles.

The same is true on the other side. We have an angle of depression of 60 degrees to the point on the bank, which we’ll call 𝑊. And we’ll label its alternate interior angle as 60 degrees. What represents the width of the river in our diagram?

The width of the river is the distance from one side to the other, from our 𝑊 to our 𝐸. We can use right triangles to help us solve this problem. The height of the aeroplane is a perpendicular distance to the river. That helps us to see that we’re dealing with two right triangles. We’ll label one of the lengths lowercase 𝑒 and the other side lowercase 𝑤.

Starting on the left, we have a right triangle. From our angle, we know the opposite side length. And we want to know the adjacent side length. This ratio, opposite and adjacent, tells us that we’ll be working with a tangent ratio. Tangent of 60 degrees equals 300 over our unknown value. We know tangent of 60 degrees equals the square root of three. We’ll substitute the square root of three for the tangent of 60 degrees in our problem. Our new equation says the square root of three equals 300 over 𝑤. I’ll multiply the right side by 𝑤. And the left side, on the right, the 𝑤s cancel out. And our new statement says the square root of three times 𝑤 equals 300.

To get 𝑤 by itself, we divide by the square root of three on the left and the right. And we find out that 𝑤 equals 300 over the square root of three. We don’t really wanna keep the square root of three in the denominator. So we rationalise by multiplying the square root of three over the square root of three. Our numerator becomes 300 times the square root of three. And in the denominator, the square root of three times the square root of three equals three.

We recognise that 300 is divisible by three. 300 divided by three is 100. Our simplified side length is then 100 times the square root of three. We plug in 100 times the square root of three into our diagram.

Looking at the other side of the triangle, we recognise something. This right triangle is a 45-45-90 triangle. And we know that, in a 45-45-90 triangle, the ratio of side lengths are always one to one to the square root of two. The two smaller sides of a 45-45-90-degree triangle are equal to each other. And that means side 𝑒 is 300 metres.

To find the width of the river, we need to combine both of these sides: 100 times the square root of three plus 300. To solve this problem, we can factor out 100: 100 times the square root of three plus three. We’ve been given an approximation for the square root of three. Substituting that in, we have 1.732 plus three. 1.732 plus three equals 4.732. If we multiply 100 times 4.732, we move the decimal two places to the right and we find that the width of the river is 473.2 metres.