Using the Pythagorean theorem, find the other leg:

\(AC=\sqrt{AB^2-BC^2}=\sqrt{25-21}=2\).

In a right triangle, \(\sin B\) is the ratio of the opposite leg to the hypotenuse:

\(\sin B=\dfrac{AC}{AB}=\dfrac{2}{5}\).

Answer: \(\dfrac{2}{5}\).

Let the second leg be \(x\). Then from the area formula for a right triangle:

\(\dfrac12\cdot 8\cdot x=60\).

\(4x=60\Rightarrow x=15\).

Now find the hypotenuse using the Pythagorean theorem:

\(c=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17\).

Answer: \(17\).

By the definition of sine:

\(\sin B=\dfrac{AC}{AB}\).

So,

\(AC=AB\cdot \sin B=10\cdot \dfrac{\sqrt{51}}{10}=\sqrt{51}\).

Now use the Pythagorean theorem:

\(BC=\sqrt{AB^2-AC^2}=\sqrt{100-51}=\sqrt{49}=7\).

Answer: \(7\).

Draw the altitude to the base. In an isosceles triangle, it bisects the base and forms a right triangle.

Let the altitude be \(h\). Then, by the definition of sine of the base angle:

\(\sin \alpha=\dfrac{h}{\sqrt{34}}=\dfrac{3}{\sqrt{34}}\).

Hence \(h=3\).

Half of the base is:

\(\sqrt{34-3^2}=\sqrt{25}=5\).

Therefore, the whole base is \(2\cdot 5=10\).

Answer: \(10\).

Find the altitude to the base using the area formula:

\(S=\dfrac12 ah\).

\(48=\dfrac12\cdot 12\cdot h\Rightarrow 48=6h\Rightarrow h=8\).

In an isosceles triangle, the altitude bisects the base, so half of the base is \(6\).

The equal side is the hypotenuse of a right triangle with legs \(6\) and \(8\):

\(b=\sqrt{6^2+8^2}=\sqrt{36+64}=10\).

Answer: \(10\).