Answer
Verified
416.1k+ views
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., ${\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AD}}} \right)^2} + {\left( {{\text{CD}}} \right)^2}$.
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., ${\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AD}}} \right)^2} + {\left( {{\text{CD}}} \right)^2}$.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE