The line joining the top of a hill to the foot of the hill makes an angle of30° with the horizontal through the foot of the hill. There is onetemple at the top of the hill and a guest house half way from the foot tothe hill . The top of the temple and the top of the guesthouse both make anelevation of 32° at the foot of the hill. If the guesthouse is 1 kmaway from the foot of the hill along the hill, find the heights of theguest house and the temple.

In the figure, GB is the hill. AG is the temple. EF is guesthouse. C is foot of the hill.

To find EF and AG.

CE = 1 km or 1000 m

In rt. △ CED, DE/CE = cos 30°

CD/1000 = √3/2

CD = 1000√3/2 = 866 m

DE/CE = sin 30°

DE/1000 = 1/2

DE = 1/2 × 1000

= 500 m

In right angled triangle CFD,

DF/CD = tan32°

DF/866 = 0.6249

DF = 0.6249 × 866

= 541.16 m

DE = 500 m.

EF = 541.16 - 500

= 41.16 m

Since E is midpoint of CG. (given halfway)

∴ CG = 2000 m.

BG/CG = sin 30°

BG/2000 = ½

BG = 1000 m

In right angled △ CBG,

CB/CG = cos 30°

CB/2000 = √3/2

∴ CB = 1732 m

In △ CDF and △ CBA

∠ CDF = ∠ CBA = 90°

∠ DCF = ∠ BCA (common)

∴ △ CDF ~ △ CBA

∴ CD/CB = DF/AB

⇒ 866/1732 = 541.16/AB

⇒ AB = 541.16 × 1732 / 866

∴ AB = 1082.32 m

AG = 1082.32-1000

= 82.32 m

∴ The height of guest house is 41 m and the height of temple is 82 m