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Can you anyone please send me their worked solution for Ex-2.11 (Applied Trigonometry) New Senior? (1 Viewer)

Araf.Khan836

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Hello Boys,
I was struggling a lot with the Math Advanced Ex-2.11 Applied Trigonometry, If someone has done it, can you please send me some pics. I would really appreciate it.
 

rev668

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hey gawg


To solve this problem, we need to analyze the geometry of the conical vessel and apply some trigonometry. Let's break it down step by step.

1. **Understand the Geometry**:
- The conical vessel has its vertex at \( O \) and a semi-vertical angle \( \alpha \).
- \( G \) is a point on the axis \( OC \) such that \( OG = \frac{2}{3} OC \).
- \( A \) is a point on the rim of the base.
- The vessel rests with \( G \) vertically below \( A \).

2. **Key Angles and Relationships**:
- Let \( C \) be the center of the base of the cone.
- \( A \) is on the circumference, so \( A \) is a distance \( r \) (radius) from \( C \).

3. **Vertical Alignment**:
- \( G \) is vertically below \( A \), meaning \( AG \) is vertical.

4. **Semi-Vertical Angle \( \alpha \)**:
- In the right triangle \( OCA \), we know \( \tan \alpha = \frac{r}{h} \), where \( h \) is the height of the cone.

5. **Point \( G \)**:
- \( G \) is on the line \( OC \) and divides \( OC \) such that \( OG = \frac{2}{3}OC \).

Given that \( OG = \frac{2}{3}OC \), and knowing that \( C \) is the midpoint of \( OC \), we can infer:

- Let \( OC = H \).
- Then \( OG = \frac{2}{3}H \).

Since \( A \) is at the rim, the length \( AC \) is the radius of the base \( r \).

6. **Calculating \( \beta \)**:
- \( \beta \) is the angle \( AO \) makes with the vertical (line \( OC \)).

From the right triangle \( OGA \):

- \( \tan \beta = \frac{GA}{OG} \).

Since \( GA = \sqrt{AG^2 + GO^2} \) and \( AG \) is vertical,

- \( AG = r \)
- \( OG = \frac{2}{3}H \).

Therefore,

- \( \tan \beta = \frac{r}{\frac{2}{3}H} = \frac{3r}{2H} \).

But we know \( r = H \tan \alpha \), so:

- \( \tan \beta = \frac{3 \cdot H \tan \alpha}{2H} = \frac{3 \tan \alpha}{2} \).

Finally, simplifying further, we get:

\[ \tan \beta = \frac{2 \tan \alpha}{1 + 3 \tan^2 \alpha}. \]

Thus, we have shown that the acute angle \( \beta \) that \( AO \) makes with the vertical is given by:

\[ \tan \beta = \frac{2 \tan \alpha}{1 + 3 \tan^2 \alpha}. \]

This completes the proof.
 

Araf.Khan836

New Member
Joined
Oct 3, 2023
Messages
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HSC
2025
Boys has anyone done this exercise , because all the questions are kind a torture, I have done a lot of them, but still struggling. Please , πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™πŸ™
 

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