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.

Question

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.

Philoid · Accepted Answer

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