Answer:
The height of the hill is 116.9 meters.
Step-by-step explanation:
The diagram depicting this problem is drawn and attached below.
From Triangle ABC
[tex]\tan 22^\circ=\dfrac{h}{150+x}\\\\h=\tan 22^\circ(150+x)[/tex]
From Triangle XBC
[tex]\tan 40^\circ =\dfrac{h}{x}\\\\h=x\tan 40^\circ[/tex]
Therefore:
[tex]h=\tan 22^\circ(150+x)=x\tan 40^\circ\\150\tan 22^\circ+x\tan 22^\circ=x\tan 40^\circ\\x\tan 40^\circ-x\tan 22^\circ=150\tan 22^\circ\\x(\tan 40^\circ-\tan 22^\circ)=150\tan 22^\circ\\x=\dfrac{150\tan 22^\circ}{\tan 40^\circ-\tan 22^\circ} \\\\x=139.30[/tex]
Therefore, the height of the hill
[tex]h=139.3\times \tan 40^\circ\\=116.9$ meters( correct to 1 d.p.)[/tex]
The height of the hill is 116.9 meters.
