To solve the problem step by step, we will follow the instructions provided in the video transcript and elaborate on each step.
### Step 1: Calculate Angle B
Given:
\[
\angle B = \sec^{-1}\left(\frac{5}{4}\right) + \csc^{-1}(\sqrt{5})
\]
Let:
\[
\sec^{-1}\left(\frac{5}{4}\right) = \alpha \quad \text{and} \quad \csc^{-1}(\sqrt{5}) = \beta
\]
From \(\sec \alpha = \frac{5}{4}\), we can create a right triangle:
- Hypotenuse = 5
- Base = 4
- Using Pythagorean theorem, the height (perpendicular) is:
\[
\sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3
\]
Thus, \(\tan \alpha = \frac{3}{4}\).
For \(\csc \beta = \sqrt{5}\):
- Hypotenuse = \(\sqrt{5}\)
- Let the base = 2 (since \(\sqrt{5}^2 - 2^2 = 1\))
- Using Pythagorean theorem, the height (perpendicular) is:
\[
\sqrt{(\sqrt{5})^2 - 2^2} = \sqrt{5 - 4} = 1
\]
Thus, \(\tan \beta = \frac{1}{2}\).
Now, we can find \(\tan(\angle B)\):
\[
\tan(\angle B) = \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\frac{3}{4} + \frac{1}{2}}{1 - \frac{3}{4} \cdot \frac{1}{2}} = \frac{\frac{3}{4} + \frac{2}{4}}{1 - \frac{3}{8}} = \frac{\frac{5}{4}}{\frac{5}{8}} = 2
\]
Thus, \(\angle B = \tan^{-1}(2)\).
### Step 2: Calculate Angle C
Given:
\[
\angle C = \csc^{-1}\left(\frac{25}{7}\right) + \cot^{-1}\left(\frac{9}{13}\right)
\]
Let:
\[
\csc \gamma = \frac{25}{7} \quad \text{and} \quad \cot \delta = \frac{9}{13}
\]
From \(\csc \gamma = \frac{25}{7}\):
- Hypotenuse = 25
- Base = 24 (since \(\sqrt{25^2 - 24^2} = 7\))
- Thus, \(\tan \gamma = \frac{7}{24}\).
For \(\cot \delta = \frac{9}{13}\):
- Thus, \(\tan \delta = \frac{13}{9}\).
Now, we can find \(\tan(\angle C)\):
\[
\tan(\angle C) = \tan(\gamma + \delta) = \frac{\tan \gamma + \tan \delta}{1 - \tan \gamma \tan \delta} = \frac{\frac{7}{24} + \frac{13}{9}}{1 - \frac{7}{24} \cdot \frac{13}{9}} = \frac{\frac{63 + 104}{216}}{1 - \frac{91}{216}} = \frac{\frac{167}{216}}{\frac{125}{216}} = \frac{167}{125}
\]
Thus, \(\angle C = \tan^{-1}\left(\frac{167}{125}\right)\).
### Step 3: Use the Sine Rule to Find Sides
Using the sine rule:
\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
\]
Given \(c = 3\) and we need to find \(a\) and \(b\).
First, we find \(\sin B\) and \(\sin C\):
- \(\sin B = \frac{2}{\sqrt{5}}\) (from \(\tan B = 2\))
- \(\sin C = \frac{3}{\sqrt{10}}\) (from \(\tan C = 3\))
Now, substituting:
\[
\frac{3}{\sin C} = \frac{a}{\sin A} \quad \text{and} \quad \frac{3}{\sin C} = \frac{b}{\sin B}
\]
From this, we can find \(a\) and \(b\):
\[
a = 3 \cdot \frac{\sin A}{\sin C} \quad \text{and} \quad b = 3 \cdot \frac{\sin B}{\sin C}
\]
### Step 4: Calculate the Distance Between Orthocenter and Centroid
The orthocenter \(H\) of the triangle is at the origin \((0, 0)\), and the centroid \(G\) can be calculated using the vertices of the triangle. The vertices can be taken as:
- \(A(0, 0)\)
- \(B(4, 0)\)
- \(C(0, 3)\)
The centroid \(G\) is given by:
\[
G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) = \left(\frac{0 + 4 + 0}{3}, \frac{0 + 0 + 3}{3}\right) = \left(\frac{4}{3}, 1\right)
\]
Now, the distance \(D\) between the orthocenter and the centroid is:
\[
D = \sqrt{\left(\frac{4}{3} - 0\right)^2 + (1 - 0)^2} = \sqrt{\left(\frac{4}{3}\right)^2 + 1^2} = \sqrt{\frac{16}{9} + 1} = \sqrt{\frac{16}{9} + \frac{9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}
\]
### Final Answer
The distance between the orthocenter and centroid of the triangle is:
\[
\boxed{\frac{5}{3}}
\]
