In the figure, given below, ABC is an isosceles triangle with BC = 8 cm and AB = AC = 5 cm. Find :
(i) sin B
(ii) tan C
(iii) `sin^(2)B + cos^(2)B`
(iv) `tan C - cot B`

Answer

To solve the problem step by step, we will analyze the isosceles triangle ABC where BC = 8 cm and AB = AC = 5 cm.

### Step 1: Draw the triangle and label the points
- Draw triangle ABC with AB = AC = 5 cm and BC = 8 cm.
- Let D be the midpoint of BC. Therefore, BD = DC = 4 cm (since BC = 8 cm).

### Step 2: Draw the altitude from A to BC
- Draw a perpendicular line from point A to line segment BC, meeting at point D. This forms two right triangles: ABD and ACD.

### Step 3: Use the Pythagorean theorem to find AD
- In triangle ABD, we can apply the Pythagorean theorem:
\[ AB^2 = AD^2 + BD^2 \]
Substituting the known values:
\[ 5^2 = AD^2 + 4^2 \]
\[ 25 = AD^2 + 16 \]
\[ AD^2 = 25 - 16 = 9 \]
\[ AD = \sqrt{9} = 3 \text{ cm} \]

### Step 4: Find sin B
- We know that:
\[ \sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AD}{AB} \]
Substituting the values:
\[ \sin B = \frac{3}{5} \]

### Step 5: Find tan C
- For angle C in triangle ACD:
\[ \tan C = \frac{\text{opposite}}{\text{adjacent}} = \frac{AD}{DC} \]
Substituting the values:
\[ \tan C = \frac{3}{4} \]

### Step 6: Find sin²B + cos²B
- We know that:
\[ \sin^2 B + \cos^2 B = 1 \]
- To find cos B, we use:
\[ \cos B = \frac{BD}{AB} = \frac{4}{5} \]
- Therefore:
\[ \sin^2 B + \cos^2 B = \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1 \]

### Step 7: Find tan C - cot B
- We know:
\[ \cot B = \frac{1}{\tan B} = \frac{BD}{AD} = \frac{4}{3} \]
- Therefore:
\[ \tan C - \cot B = \frac{3}{4} - \frac{4}{3} \]
- To subtract these fractions, we find a common denominator (12):
\[ \tan C - \cot B = \frac{3 \times 3}{4 \times 3} - \frac{4 \times 4}{3 \times 4} = \frac{9}{12} - \frac{16}{12} = \frac{9 - 16}{12} = \frac{-7}{12} \]

### Final Answers
(i) \( \sin B = \frac{3}{5} \)
(ii) \( \tan C = \frac{3}{4} \)
(iii) \( \sin^2 B + \cos^2 B = 1 \)
(iv) \( \tan C - \cot B = -\frac{7}{12} \)