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: \(10\).

Let the given leg be \(4\), and let it be opposite the angle \(\alpha\). Then, by the definition of tangent:

\(\tan \alpha=\dfrac{4}{x}=\dfrac{2}{\sqrt5}\),

where \(x\) is the other leg.

Hence:

\(4\sqrt5=2x\Rightarrow x=2\sqrt5\).

Now find the hypotenuse:

\(c=\sqrt{4^2+(2\sqrt5)^2}=\sqrt{16+20}=\sqrt{36}=6\).

Answer: \(6\).

Let the altitude be drawn to the side that it divides into segments \(5\) and \(9\). Then this side equals:

\(5+9=14\).

This gives two right triangles. Their hypotenuses are the other two sides of the triangle:

\(a=\sqrt{12^2+5^2}=\sqrt{144+25}=13\),

\(b=\sqrt{12^2+9^2}=\sqrt{144+81}=15\).

Hence the perimeter is:

\(P=13+15+14=42\).

Answer: \(42\).

The base of the triangle is \(4\), so half of it is \(2\).

Draw the altitude to the base. Then we get a right triangle whose hypotenuse is \(2\sqrt{10}\), and one leg is \(2\).

The altitude is:

\(h=\sqrt{(2\sqrt{10})^2-2^2}=\sqrt{40-4}=6\).

The area is:

\(S=\dfrac12\cdot 4\cdot 6=12\).

Answer: \(12\).

Using the Pythagorean theorem, find the other leg:

\(AC=\sqrt{AB^2-BC^2}=\sqrt{64-28}=\sqrt{36}=6\).

In a right triangle, \(\cos A\) is the ratio of the adjacent leg to the hypotenuse:

\(\cos A=\dfrac{AC}{AB}=\dfrac{6}{8}=\dfrac{3}{4}\).

Answer: \(\dfrac{3}{4}\).

Let the second leg be \(x\). Then:

\(\dfrac12\cdot 12\cdot x=96\).

\(6x=96\Rightarrow x=16\).

By the Pythagorean theorem:

\(c=\sqrt{12^2+16^2}=\sqrt{144+256}=\sqrt{400}=20\).

Answer: \(20\).

By the definition of cosine:

\(\cos A=\dfrac{AC}{AB}\).

Then

\(AC=AB\cdot \cos A=5\cdot \dfrac{2\sqrt6}{5}=2\sqrt6\).

Now use the Pythagorean theorem:

\(BC=\sqrt{AB^2-AC^2}=\sqrt{25-24}=1\).

Answer: \(1\).

Draw the altitude to the base. It bisects the base, so half of the base is \(2\).

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

\(\tan \alpha=\dfrac{h}{2}=\dfrac{\sqrt5}{2}\).

Hence, \(h=\sqrt5\).

The equal side is the hypotenuse of a right triangle:

\(b=\sqrt{2^2+(\sqrt5)^2}=\sqrt{4+5}=3\).

Answer: \(3\).

Using the area formula:

\(108=\dfrac12\cdot a\cdot 9\).

\(108=\dfrac{9a}{2}\Rightarrow 216=9a\Rightarrow a=24\).

Half of the base is \(12\).

The equal side is the hypotenuse of a right triangle with legs \(12\) and \(9\):

\(b=\sqrt{12^2+9^2}=\sqrt{144+81}=\sqrt{225}=15\).

Answer: \(15\).

Let the given leg of length \(3\) be adjacent to the angle \(\alpha\). Then, by the definition of tangent:

\(\tan \alpha=\dfrac{x}{3}=\dfrac{2\sqrt{10}}{3}\),

where \(x\) is the other leg.

Hence:

\(x=2\sqrt{10}\).

Now find the hypotenuse using the Pythagorean theorem:

\(c=\sqrt{3^2+(2\sqrt{10})^2}=\sqrt{9+40}=\sqrt{49}=7\).

Answer: \(7\).

The side to which the altitude is drawn equals:

\(8+20=28\).

This gives two right triangles. Their hypotenuses are the other two sides of the original triangle:

\(a=\sqrt{15^2+8^2}=\sqrt{225+64}=17\),

\(b=\sqrt{15^2+20^2}=\sqrt{225+400}=25\).

Therefore, the perimeter is:

\(P=17+25+28=70\).

Answer: \(70\).

The base is \(6\), so half of it is \(3\).

Draw the altitude to the base. It forms a right triangle with hypotenuse \(3\sqrt5\) and one leg \(3\).

The altitude is:

\(h=\sqrt{(3\sqrt5)^2-3^2}=\sqrt{45-9}=6\).

The area of the triangle is:

\(S=\dfrac12\cdot 6\cdot 6=18\).

Answer: \(18\).