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.

https://www.studyrankersonline.com/?qa=blob&qa_blobid=11286063250835042824

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

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