Given a triangle STD, ST=DV, <TSD=48°, <DVS=18°. Find X

[tato1982]

Given a triangle STD, ST=DV, <TSD=48°, <DVS=18°. Find an angle X.

[Guest]

first, [tex]\angle TDS=x+{18} ^\circ[/tex] by the exterior angle property. Now, using the sine rule in [tex]\triangle TSD[/tex],

[tex]\frac{TS}{\sin{\angle TDS}}=\frac{TD}{\sin{\angle TSD}} \Longrightarrow \frac{TS}{\sin{(x+{18} ^\circ )}}=\frac{TD}{\sin{{48} ^\circ }}\Longrightarrow TD=\sin{{48} ^\circ }\left\{\frac{TS}{\sin{(x+{18} ^\circ )}}\right\}\ldots(1)[/tex]

Now, similarly in [tex]\triangle TDV[/tex],

[tex]\frac{DV}{\sin{\angle DTV}}=\frac{TD}{\sin{\angle TVD}} \Longrightarrow \frac{DV}{\sin{x}}=\frac{TD}{\sin{{18} ^\circ }}\Longrightarrow TD=\sin{{18} ^\circ }\left(\frac{DV}{\sin{x}}\right)\ldots(2)[/tex]

from [tex](1)[/tex] and [tex](2)[/tex] we get,

[tex]\sin{{48} ^\circ }\left\{\frac{TS}{\sin{(x+{18} ^\circ )}}\right\}=\sin{{18} ^\circ }\left(\frac{DV}{\sin{x}}\right)\Longrightarrow\frac{\sin{{48} ^\circ }}{\sin{(x+{18} ^\circ )}}=\frac{\sin{{18} ^\circ }}{\sin{x}}[/tex] because [tex]ST=DV[/tex], they can be cancelled.

I hope you can solve it from here. just use the expansion formula for [tex]\sin{(A+B)}[/tex] and solve for [tex]\sin{x}[/tex].

[Guest]

first, [tex]\angle TDS=x+{18} ^\circ[/tex] by the exterior angle property. Now, using the sine rule in [tex]\triangle TSD[/tex],

[tex]\frac{TS}{\sin{\angle TDS}}=\frac{TD}{\sin{\angle TSD}} \Longrightarrow \frac{TS}{\sin{(x+{18} ^\circ )}}=\frac{TD}{\sin{{48} ^\circ }}\Longrightarrow TD=\sin{{48} ^\circ }\left\{\frac{TS}{\sin{(x+{18} ^\circ )}}\right\}\ldots(1)[/tex]

Now, similarly in [tex]\triangle TDV[/tex],

[tex]\frac{DV}{\sin{\angle DTV}}=\frac{TD}{\sin{\angle TVD}} \Longrightarrow \frac{DV}{\sin{x}}=\frac{TD}{\sin{{18} ^\circ }}\Longrightarrow TD=\sin{{18} ^\circ }\left(\frac{DV}{\sin{x}}\right)\ldots(2)[/tex]

from [tex](1)[/tex] and [tex](2)[/tex] we get,

[tex]\sin{{48} ^\circ }\left\{\frac{TS}{\sin{(x+{18} ^\circ )}}\right\}=\sin{{18} ^\circ }\left(\frac{DV}{\sin{x}}\right)\Longrightarrow\frac{\sin{{48} ^\circ }}{\sin{(x+{18} ^\circ )}}=\frac{\sin{{18} ^\circ }}{\sin{x}}[/tex] because [tex]ST=DV[/tex], they can be cancelled.

I hope you can solve it from here. just use the expansion formula for [tex]\sin{(A+B)}[/tex] and solve for [tex]\sin{x}[/tex].

continuing from here,

[tex]\Longrightarrow\sin{{48} ^\circ }\sin{x}=\sin{{18} ^\circ }\sin{(x+{18} ^\circ )}\Longrightarrow\sin{{48} ^\circ }\sin{x}=\sin{{18} ^\circ }(\sin{x}\cos{{18}^\circ}+\cos{x}\sin{{18}^\circ})[/tex]
[tex]\Longrightarrow\sin{x}(\sin{{48}^\circ}-\sin{{18}^\circ}\cos{{18}^\circ})={\sin}^2{18}^\circ\cos{x}[/tex]
[tex]\Longrightarrow\tan{x}=\frac{{\sin}^2{18}^\circ}{\sin{{48}^\circ}-\sin{{18}^\circ}\cos{{18}^\circ}}[/tex]
[tex]\Longrightarrow x=12^\circ[/tex]