We are given a right-angled triangle.
The given angle is 50°40'
Let us convert the minute part to degrees
[tex]\begin{gathered} \frac{40^{\prime}}{60}=0.67\degree \\ 50\degree+0.67\degree=50.67\degree \end{gathered}[/tex]
With respect to the angle 50.67°, the adjacent side is 17.2' and the opposite side is b.
Recall from the trignonometric ratios,
[tex]\tan \theta=\frac{opposite}{adjacent}[/tex]
For the given case, we have
θ = 50.67°
Adjacent = 17.2'
Opposite side = b
Let us substitute these values into the above formula to find the value of b.
[tex]\begin{gathered} \tan (50.67\degree)=\frac{b}{17.2} \\ b=\tan (50.67\degree)\cdot17.2 \\ b=1.2205\cdot17.2 \\ b=21.0^{\prime} \end{gathered}[/tex]
Therefore, the value of b is 21.0'
