Given:
The figure of triangle ABC.
The area of the triangle ABC is D.
[tex]m\angle B=\sin ^{-1}(\dfrac{m}{n})[/tex]
To find:
The value of m and n in the given expression.
Solution:
Let h be the height of the triangle ABC.
Area of a triangle is:
[tex]Area=\dfrac{1}{2}\times base\times h[/tex]
Where, b is the base and h is the height of the triangle.
[tex]Area=\dfrac{1}{2}\times a\times h[/tex]
The area of the triangle ABC is D.
[tex]D=\dfrac{1}{2}\times a\times h[/tex]
[tex]2D=ah[/tex]
[tex]\dfrac{2D}{a}=h[/tex] ...(i)
In a right angle triangle,
[tex]\sin \theta =\dfrac{Perpendicular}{Hypotenuse}[/tex]
[tex]\sin B =\dfrac{h}{c}[/tex]
[tex]\sin B =\dfrac{1}{c}\times \dfrac{2D}{a}[/tex] [Using (i)]
[tex]\sin B =\dfrac{2D}{ac}[/tex]
[tex]m\angle B =\sin ^{-1}\dfrac{2D}{ac}[/tex] ...(ii)
We have,
[tex]m\angle B=\sin ^{-1}(\dfrac{m}{n})[/tex] ...(iii)
On comparing (ii) and (iii), we get
[tex]m=2D[/tex]
[tex]n=ac[/tex]
Therefore, the required values are [tex]m=2D, n=ac[/tex].
