Question

From a point on a horizontal plane, the elevation of the top of the hill is 45°. The elevation
becomes 75° after walking a distance of 500 m up a slope inclined at an angle of 15° to the
horizon. The height of the hill is:
a) 500V6 m
by 50073 m
c) 250V6 m
d) 250V3 m saber
RPS​

Answer 1

The height of the hill is 612.371 m or option (c) 250√6 m.

Step-by-step explanation:

Referring to the attached figure below, let’s make some assumptions

AC = the height of the hill

∠AEC = 45° = the angle of elevation of the top of the hill

∠FED = 15° = angle of inclination of the slope to the horizon

EF = 500 m = distance of the slope  

∠AFB = 75° = angle of elevation after walking 500 m up the slope

In ∆ FDE, applying trigonometric ratios of a triangle, we get

sin 15° = perpendicular/hypotenuse

⇒ sin 15° = FD/EF

⇒ 0.2588 = FD/500

⇒ FD = 129.40 m …… (i)

Also,

cos 15° = base/hypotenuse

⇒ cos 15° = ED/EF

⇒ 0.9659 = ED/500

⇒ ED = 482.95 m …… (ii)

In ∆ ACE, applying trigonometric ratios of a triangle, we get

tan 45° = perpendicular/base

⇒ tan 45° = AC/EC

⇒ 1 = AC/EC

⇒ AC = EC ...... (iii)

Now, in ΔABC, applying trigonometric ratios of a triangle, we get

tan 75° = perpendicular/base

⇒ tan 75° = AB/BF

⇒ 3.732 = [AC - BC] / [CD] ....... [since BF = CD & FD = BC as shown in the figure]

⇒ 3.732 = [AC - BC] / [EC - ED]

substituting from (i), (ii) & (iii)

⇒ 3.732 = [AC - 129.40] / [AC - 482.96]

⇒ 3.732AC - 1802.40 = AC - 129.40

⇒ 3.732AC - 1802.40 = AC - 129.40

⇒ 3.732AC - AC = 1802.40 - 129.40

⇒ 2.732AC = 1673

⇒ AC = $\frac{1673}{2.732}$

⇒ AC = 612.371 m or 250√6 m (as given in option) ← height of the hill

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