In the rectangle, diagonal \(AC\) is the hypotenuse of right triangle \(ABC\).

By Pythagoras' theorem:

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

For angle \(\angle BAC\), the opposite side is \(BC\) and the hypotenuse is \(AC\).

So, \(\sin \angle BAC=\dfrac{BC}{AC}=\dfrac{8}{10}=\dfrac{4}{5}\).

First find the other leg:

\(AC=\sqrt{15^2-9^2}=\sqrt{225-81}=12\).

For angle \(B\), the opposite side is \(AC\) and the hypotenuse is \(AB\).

So, \(\sin B=\dfrac{AC}{AB}=\dfrac{12}{15}=\dfrac{4}{5}\).

By the definition of sine:

\(\sin B=\dfrac{AC}{AB}=\dfrac{5}{13}\).

So \(AC=26\cdot\dfrac{5}{13}=10\text{ cm}\).

Now by Pythagoras' theorem:

\(BC=\sqrt{26^2-10^2}=\sqrt{676-100}=24\text{ cm}\).

In right triangle \(ABC\):

\(AC=\sqrt{24^2+7^2}=\sqrt{576+49}=25\).

For angle \(\angle BAC\), the adjacent leg is \(AB\) and the hypotenuse is \(AC\).

Therefore, \(\cos \angle BAC=\dfrac{AB}{AC}=\dfrac{24}{25}\).

Find the other leg:

\(BC=\sqrt{25^2-24^2}=\sqrt{625-576}=7\).

For angle \(B\), the adjacent leg is \(BC\) and the hypotenuse is \(AB\).

So, \(\cos B=\dfrac{BC}{AB}=\dfrac{7}{25}\).

By the definition of cosine:

\(\cos B=\dfrac{BC}{AB}=\dfrac{7}{25}\).

So \(BC=25\cdot\dfrac{7}{25}=7\text{ cm}\).

Now by Pythagoras' theorem:

\(AC=\sqrt{25^2-7^2}=\sqrt{625-49}=24\text{ cm}\).

Consider right triangle \(ABD\).

The diagonal is \(BD=\sqrt{12^2+16^2}=\sqrt{144+256}=20\).

For angle \(\angle ABD\), the opposite leg is \(AD\) and the hypotenuse is \(BD\).

Therefore, \(\sin \angle ABD=\dfrac{AD}{BD}=\dfrac{16}{20}=\dfrac{4}{5}\).

Find the leg \(BC\):

\(BC=\sqrt{30^2-24^2}=\sqrt{900-576}=18\).

For angle \(A\), the opposite leg is \(BC\) and the adjacent leg is \(AC\).

So, \(\tan A=\dfrac{BC}{AC}=\dfrac{18}{24}=\dfrac{3}{4}\).

By the definition of tangent:

\(\tan A=\dfrac{BC}{AC}=\dfrac{4}{3}\).

Since \(BC=12\), we have \(\dfrac{12}{AC}=\dfrac{4}{3}\).

Hence \(AC=9\text{ cm}\).

Then the hypotenuse is:

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

In right triangle \(ABC\):

\(AB=\sqrt{25^2-20^2}=\sqrt{625-400}=15\).

For angle \(\angle BAC\), the adjacent leg is \(AB\) and the hypotenuse is \(AC\).

Therefore, \(\cos \angle BAC=\dfrac{15}{25}=\dfrac{3}{5}\).

Find the leg \(AC\):

\(AC=\sqrt{10^2-6^2}=\sqrt{100-36}=8\).

For angle \(A\), the adjacent leg is \(AC\) and the hypotenuse is \(AB\).

So, \(\cos A=\dfrac{AC}{AB}=\dfrac{8}{10}=\dfrac{4}{5}\).

By the definition of cosine:

\(\cos A=\dfrac{AC}{AB}=\dfrac{12}{13}\).

So \(AC=26\cdot\dfrac{12}{13}=24\text{ cm}\).

By Pythagoras' theorem:

\(BC=\sqrt{26^2-24^2}=\sqrt{676-576}=10\text{ cm}\).

In right triangle \(ABC\):

\(BC=\sqrt{30^2-18^2}=\sqrt{900-324}=24\).

For angle \(\angle BAC\), the opposite leg is \(BC\) and the hypotenuse is \(AC\).

Therefore, \(\sin \angle BAC=\dfrac{24}{30}=\dfrac{4}{5}\).

Find the leg \(BC\):

\(BC=\sqrt{20^2-12^2}=\sqrt{400-144}=16\).

For angle \(A\), the opposite leg is \(BC\) and the hypotenuse is \(AB\).

So, \(\sin A=\dfrac{16}{20}=\dfrac{4}{5}\).

By the definition of sine:

\(\sin B=\dfrac{AC}{AB}=\dfrac{8}{17}\).

So \(AC=34\cdot\dfrac{8}{17}=16\text{ cm}\).

Now by Pythagoras' theorem:

\(BC=\sqrt{34^2-16^2}=\sqrt{1156-256}=30\text{ cm}\).

Consider right triangle \(BCD\).

By Pythagoras' theorem:

\(BC=\sqrt{15^2-12^2}=\sqrt{225-144}=9\).

For angle \(\angle BDC\), the opposite leg is \(BC\) and the adjacent leg is \(CD\).

Therefore, \(\tan \angle BDC=\dfrac{BC}{CD}=\dfrac{9}{12}=\dfrac{3}{4}\).