## Trigonometry (11th Edition) Clone

Let $d$ be the horizontal distance from the building to the tower. Let $x$ be the height of the tower. We can use the law of sines and the angle at the base of the building to find an expression for $d$: $\frac{d}{sin(90^{\circ}-29^{\circ}30')} = \frac{x}{sin(29^{\circ}30')}$ $d = \frac{x~sin(60^{\circ}30')}{sin(29^{\circ}30')}$ We can use the law of sines and the angle on the roof of the building to find an expression for $d$: $\frac{d}{sin(90^{\circ}-15^{\circ}20')} = \frac{x-45}{sin(15^{\circ}20')}$ $d = \frac{(x-45)~sin(74^{\circ}40')}{sin(15^{\circ}20')}$ We can equate the two expressions for $d$ and solve for $x$: $\frac{x~sin(60^{\circ}30')}{sin(29^{\circ}30')} = \frac{(x-45)~sin(74^{\circ}40')}{sin(15^{\circ}20')}$ $x~sin(60^{\circ}30')~sin(15^{\circ}20') = (x-45)~sin(74^{\circ}40')~sin(29^{\circ}30')$ $0.2301469~x = 0.474896~(x-45)$ $0.24475~x = 21.37$ $x = \frac{21.37}{0.24475}$ $x = 87.3~ft$ The height of the tower is 87.3 ft