From the given figure, we can separate the diagram into two right-angled triagle as shown below
The figure is divided into right-angle trainle DAEand BAD. Please note that
[tex]\begin{gathered} CD=x(\text{given)} \\ BE=y(given) \\ BC=40m(\text{given)} \\ CD=AB=x(opposite\text{ sides of a rectangle are equal)} \\ BC=AD=40m(\text{opposite sides of a rectangle are equal)} \\ AE=BE-AB=y-x \end{gathered}[/tex]
From triangle BAD, using trigonometry, we can solve for x
[tex]\begin{gathered} \tan 26^0=\frac{AB}{AD} \\ \tan 26^0=\frac{x}{40} \\ x=40\times\tan 26^0 \\ x=40\times0.4877 \\ x=19.5093m \end{gathered}[/tex]
From triangle DAE, using trigonometry, we can solve for y
[tex]\begin{gathered} \tan 37^0=\frac{AE}{AD} \\ \tan 37^0=\frac{y-x}{40} \\ y-x=40\times\tan 37^0 \\ y-x=40\times0.7536 \\ y-x=30.1422 \\ y=30.1422+x \end{gathered}[/tex][tex]\begin{gathered} x=19.5093 \\ y=30.1422+19.5093 \\ y=49.6514m \\ y=49.7m(\text{nearest tenth)} \end{gathered}[/tex]
Hence, the height of the taller building is 49.7m
