To solve the problwm we must know about trignometric functions.
[tex]Sin \theta=\dfrac{Perpendicular}{Hypotenuse}[/tex]
[tex]Cos \theta=\dfrac{Base}{Hypotenuse}[/tex]
[tex]Tan \theta=\dfrac{Perpendicular}{Base}[/tex]
where perpendicular is the side of the triangle which is opposite to the angle, and the hypotenuse is the longest side of the triangle which is opposite to the 90° angle.
The area of the triangle is 46.672 cm².
Given to us
Assumption
Let the altitude(QS) be h, and the base of the triangle(PR) be b.
A.) The altitude of the triangle
In ΔPQS
[tex]Sin(55^o\ 30') = \dfrac{Perpendicular}{Hypotenuse}\\\\Sin(55^o\ 30') = \dfrac{QS}{PQ}\\\\Sin(55^o\ 30') = \dfrac{h}{10}\\\\h = Sin(55^o\ 30') \times 10\\\\h = 8.24[/tex]
Hence, the altitude of the isosceles triangle is 8.24 cm.
B.) The length of the base
In ΔPQS
[tex]Cos(55^o\ 30') = \dfrac{Base}{Hypotenuse}\\\\Cos(55^o\ 30') = \dfrac{PS}{PQ}\\\\Cos(55^o\ 30') = \dfrac{\dfrac{b}{2}}{10}\\\\\dfrac{b}{2} = cos(55^o\ 30') \times 10\\\\{b} = cos(55^o\ 30') \times 10\times 2\\\\b = 11.33[/tex]
Hence, the base of the isosceles triangle is 11.33 cm.
C.) The area of the triangle
[tex]Area\ \triangle = \dfrac{1}{2}\times Height(altitude)\times base[/tex]
[tex]Area\ \triangle PQR = \dfrac{1}{2}\times h\times b[/tex]
[tex]= \dfrac{1}{2}\times 8.24\times 11.33\\\\=46.672\ cm^2[/tex]
Hence, the area of the triangle is 46.672 cm².
Learn more about Trignometric functions:
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