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IPMAT Indore 2024 PYQsMultiple Choice (Quants). Free, no login required.

Geometry

past year questions, IPMAT Indore 2024

Trigonometry

past year questions, Geometry, IPMAT Indore 2024

Easy

The angle of elevation of the top of a pole from a point A on the ground is 30°. The angle of elevation changes to 45°, after moving 20 meters towards the base of the pole. Then the height of the pole, in meters, is

Solution

✅ Correct Option: 4

Let the height of the pole be hh and the distance from the second point (where angle =45°= 45°) to the base of the pole be dd.

From the second point (after walking 20 m toward the pole):

tan⁡45°=hd\tan 45° = \dfrac{h}{d}

1=hd1 = \dfrac{h}{d}

d=hd = h

💡 Since tan⁡45°=1\tan 45° = 1, the height and base distance from this point are equal.

From point A, the total base distance =20+d=20+h= 20 + d = 20 + h

tan⁡30°=h20+h\tan 30° = \dfrac{h}{20 + h}

13=h20+h\dfrac{1}{\sqrt{3}} = \dfrac{h}{20 + h}

Cross-multiplying:

20+h=3⋅h20 + h = \sqrt{3} \cdot h

20=3⋅h−h20 = \sqrt{3} \cdot h - h

20=h(3−1)20 = h(\sqrt{3} - 1)

h=203−1h = \dfrac{20}{\sqrt{3} - 1}

Rationalizing the denominator (multiplying top and bottom by 3+1\sqrt{3} + 1 to remove the square root from the denominator):

h=203−1×3+13+1h = \dfrac{20}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1}

h=20(3+1)(3)2−(1)2h = \dfrac{20(\sqrt{3} + 1)}{(\sqrt{3})^2 - (1)^2}

h=20(3+1)3−1h = \dfrac{20(\sqrt{3} + 1)}{3 - 1}

h=20(3+1)2h = \dfrac{20(\sqrt{3} + 1)}{2}

h=10(3+1) meters\boxed{h = 10(\sqrt{3} + 1) \text{ meters}}

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