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Hint-Hint- Here, we will proceed by drawing the figure corresponding to the problem statement and then, we will use the identity to find cosB which will eventually give the value of the length AB.

__Complete step-by-step answer:__

In $\vartriangle $ABC, $\sin {\text{B}} = 0.8$, BD = 9 cm and $\tan {\text{C}} = 1$

AD is the perpendicular drawn from point A to side BC of the triangle ABC which will divide the triangle ABC into two right angled triangles with right angle at vertex D.

As we know that ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1$

For angle B using the above identity, we can write

${\left( {\sin {\text{B}}} \right)^2} + {\left( {\cos {\text{B}}} \right)^2} = 1$

By substituting $\sin {\text{B}} = 0.8$ in the above equation, we get

$

\Rightarrow {\left( {0.8} \right)^2} + {\left( {\cos {\text{B}}} \right)^2} = 1 \\

\Rightarrow {\left( {\cos {\text{B}}} \right)^2} = 1 - {\left( {0.8} \right)^2} \\

\Rightarrow {\left( {\cos {\text{B}}} \right)^2} = 1 - 0.64 \\

\Rightarrow {\left( {\cos {\text{B}}} \right)^2} = 0.36 \\

\Rightarrow \cos {\text{B}} = \pm \sqrt {0.36} \\

\Rightarrow \cos {\text{B}} = \pm 0.6 \\

$

In any right angled triangle, according to the definition of cosine trigonometric function, we have

$\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$

Using the above formula, we can write

$

\cos {\text{B}} = \dfrac{{{\text{BD}}}}{{{\text{AB}}}} \\

\Rightarrow {\text{AB}} = \dfrac{{{\text{BD}}}}{{\cos {\text{B}}}} \\

$

By substituting BD = 9 cm and $\cos {\text{B}} = \pm 0.6$ in the above equation, we get

$ \Rightarrow {\text{AB}} = \dfrac{{\text{9}}}{{ \pm 0.6}}$

Here, we will be considering only the positive value of cosine of angle B because the length of the side of any triangle is always positive. So, cosB = -0.6 is neglected.

$

\Rightarrow {\text{AB}} = \dfrac{{\text{9}}}{{0.6}} = \dfrac{{90}}{6} \\

\Rightarrow {\text{AB}} = 15{\text{ cm}} \\

$

Therefore, the length of side AB of the triangle is 15 cm.

Note- In this particular problem, we can also find the length AD using the Pythagoras Theorem in triangle ABD i.e., ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2} \Rightarrow {\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AD}}} \right)^2} + {\left( {{\text{BD}}} \right)^2}$. Also, using $\tan {\text{C}} = \dfrac{{{\text{AD}}}}{{{\text{CD}}}}$ we will get the length CD and for the length AC we will use Pythagoras Theorem in triangle ACD i.e., ${\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AD}}} \right)^2} + {\left( {{\text{CD}}} \right)^2}$.

In $\vartriangle $ABC, $\sin {\text{B}} = 0.8$, BD = 9 cm and $\tan {\text{C}} = 1$

AD is the perpendicular drawn from point A to side BC of the triangle ABC which will divide the triangle ABC into two right angled triangles with right angle at vertex D.

As we know that ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1$

For angle B using the above identity, we can write

${\left( {\sin {\text{B}}} \right)^2} + {\left( {\cos {\text{B}}} \right)^2} = 1$

By substituting $\sin {\text{B}} = 0.8$ in the above equation, we get

$

\Rightarrow {\left( {0.8} \right)^2} + {\left( {\cos {\text{B}}} \right)^2} = 1 \\

\Rightarrow {\left( {\cos {\text{B}}} \right)^2} = 1 - {\left( {0.8} \right)^2} \\

\Rightarrow {\left( {\cos {\text{B}}} \right)^2} = 1 - 0.64 \\

\Rightarrow {\left( {\cos {\text{B}}} \right)^2} = 0.36 \\

\Rightarrow \cos {\text{B}} = \pm \sqrt {0.36} \\

\Rightarrow \cos {\text{B}} = \pm 0.6 \\

$

In any right angled triangle, according to the definition of cosine trigonometric function, we have

$\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$

Using the above formula, we can write

$

\cos {\text{B}} = \dfrac{{{\text{BD}}}}{{{\text{AB}}}} \\

\Rightarrow {\text{AB}} = \dfrac{{{\text{BD}}}}{{\cos {\text{B}}}} \\

$

By substituting BD = 9 cm and $\cos {\text{B}} = \pm 0.6$ in the above equation, we get

$ \Rightarrow {\text{AB}} = \dfrac{{\text{9}}}{{ \pm 0.6}}$

Here, we will be considering only the positive value of cosine of angle B because the length of the side of any triangle is always positive. So, cosB = -0.6 is neglected.

$

\Rightarrow {\text{AB}} = \dfrac{{\text{9}}}{{0.6}} = \dfrac{{90}}{6} \\

\Rightarrow {\text{AB}} = 15{\text{ cm}} \\

$

Therefore, the length of side AB of the triangle is 15 cm.

Note- In this particular problem, we can also find the length AD using the Pythagoras Theorem in triangle ABD i.e., ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2} \Rightarrow {\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AD}}} \right)^2} + {\left( {{\text{BD}}} \right)^2}$. Also, using $\tan {\text{C}} = \dfrac{{{\text{AD}}}}{{{\text{CD}}}}$ we will get the length CD and for the length AC we will use Pythagoras Theorem in triangle ACD i.e., ${\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AD}}} \right)^2} + {\left( {{\text{CD}}} \right)^2}$.

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