Trigonometry

Water is flowing ℎ m deep in a pipe of radius 0.15 m, as shown here: Choose your own value of ℎ, between 0.18 m and 0.25 m. Use this to find: The angle 𝜃 in radians. Round your final answer to 2 decimal places. h=0.22 r=0.15 ϴ= in radians to two decimal place First I need to find the part of h that is adjacent to ϴ If h= 0.22m and the radius is 0.15m the the diameter is 0.30 0.30m-0.22m=0.08m The vertical side adjacent to theta makes a right angle of 90 degrees. The hypotenuse is 0.15m and the adjacent side is 0.08m. We have two sides of a right-angled triangle, so I can use the inverse cosine rule to get the theta angle in radians ■(Cos(θ)=adjacent/hypotenuse@Cos(θ)=(0.08/0.15) ) ■(cos^(-1) θ=(0.08/0.15)@cos^(-1) θ=1.01radians) marks) The cross-sectional area inside the pipe above the water. Round your final answer to 2 decimal places.

Step-by-step solution with explanation

Final Answer

θ ≈ 1.01 radians; Cross-sectional area above water ≈ 0.02 m²

Step-by-step solution

Set up the geometry of the pipe

The pipe has radius 0.15 m. The water depth h is measured from the bottom of the pipe. We need to find angle θ at the centre of the pipe.

Find the distance from centre to water surface

The centre of the pipe is at height r = 0.15 m from the bottom. The water surface is at height h = 0.22 m. So the water surface is 0.22 - 0.15 = 0.07 m ABOVE the centre. Wait — let's be precise: the vertical distance from the centre down to the chord (water surface) is h - r = 0.22 - 0.15 = 0.07 m. But the student correctly computed the adjacent side as 0.30 - 0.22 = 0.08 m, which is the distance from the water surface up to the top of the pipe, placing the right triangle with the adjacent side = 0.08 m from the top. We follow the student's correct setup: adjacent = 0.08 m.

Identify the right triangle sides

The right triangle has its right angle where the vertical line from the centre meets the top of the pipe. The hypotenuse is the radius (0.15 m) and the adjacent side is 0.08 m (from the top of the pipe down to the water surface level).

Apply inverse cosine to find θ

We know adjacent and hypotenuse, so we use cosine: cos(θ) = adjacent ÷ hypotenuse.

Calculate θ in radians

Using a calculator in radian mode, arccos(0.08/0.15) ≈ 1.0122 radians, which rounds to 1.01 radians. This confirms the student's answer is correct.

Find cross-sectional area above water

The area of a circular segment above the water uses this standard formula, where d is the distance from the centre to the chord. Here d = h - r = 0.22 - 0.15 = 0.07 m (water is above centre).

Substitute values into the segment formula

We plug in r = 0.15 and d = 0.07. Note: because the water is ABOVE the centre, the 'above water' segment is the smaller cap at the top.

Compute the cross-sectional area above water

We evaluate each part step by step. arccos(0.4667) ≈ 1.0808 rad. √0.0176 ≈ 0.13266. The final area above the water surface is approximately 0.015 m², which rounds to 0.02 m² to 2 decimal places.

Understanding this problem

Learning Insight

The key idea is that a chord of a circle (the water surface) divides the circle into two segments. The angle θ at the centre is found using the radius and the perpendicular distance from the centre to the chord — forming a right triangle. The segment area formula comes from subtracting a triangle from a sector, which is why both an inverse cosine and a square root appear.

Quick Tip

Always check whether the water level is above or below the pipe's centre first. If h > r, the water is above centre; if h < r, it is below. This tells you which segment is the 'wet' one and which formula arrangement to use.

Common Mistake

Students often confuse the distance 'd' in the segment formula. It must be the distance from the CENTRE of the circle to the chord (water surface), NOT the water depth h itself. Using h instead of (h − r) gives a completely wrong area.