d:[["$","title","0",{"children":"IIM Kozhikode - If a pole 6 m high casts a shadow 2sqrt(3) m long on the ground, then the sun's elevation is | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Given Information: - Po<= height = 6 m - Shadow <=ngth = 2sqrt(3) m - Find: Sun's e<=vation ang<=

When the sun shines on a vertical po<=, it creates a right triang<= where the po<= is the vertical side (opposite to the e<=vation ang<=) and the shadow is the horizontal side (adjacent to the e<=vation ang<=).

Since we have the opposite side (po<=) and adjacent side (shadow), we use the tan>=nt ratio: (e<=vation ang<=) = oppositeadjacent = heightshadow

Let = sun's e<=vation ang<= () = 62sqrt(3)

Simplifying the fraction: () = 62sqrt(3) = 3sqrt(3) Rationalizing the denominator: () = 3sqrt(3) sqrt(3)sqrt(3) = 3sqrt(3)3 = sqrt(3)

From standard trigonometric values: (60°) = sqrt(3) Therefore: = 60°

The sun's e<=vation is 60°."}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"IIM-K (BMSAT),QA,Geometry,Trigonometry"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"IIM Kozhikode - If a pole 6 m high casts a shadow 2sqrt(3) m long on the ground, then the sun's elevation is | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Given Information: - Po<= height = 6 m - Shadow <=ngth = 2sqrt(3) m - Find: Sun's e<=vation ang<=

Since we have the opposite side (po<=) and adjacent side (shadow), we use the tan>=nt ratio: (e<=vation ang<=) = oppositeadjacent = heightshadow

Let = sun's e<=vation ang<= () = 62sqrt(3)

Simplifying the fraction: () = 62sqrt(3) = 3sqrt(3) Rationalizing the denominator: () = 3sqrt(3) sqrt(3)sqrt(3) = 3sqrt(3)3 = sqrt(3)

From standard trigonometric values: (60°) = sqrt(3) Therefore: = 60°

The sun's e<=vation is 60°."}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/iimk-bmsat/2025/QA/36"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | IIM Kozhikode Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"IIM Kozhikode - If a pole 6 m high casts a shadow 2sqrt(3) m long on the ground, then the sun's elevation is | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Given Information: - Po<= height = 6 m - Shadow <=ngth = 2sqrt(3) m - Find: Sun's e<=vation ang<=

Since we have the opposite side (po<=) and adjacent side (shadow), we use the tan>=nt ratio: (e<=vation ang<=) = oppositeadjacent = heightshadow

Let = sun's e<=vation ang<= () = 62sqrt(3)

Simplifying the fraction: () = 62sqrt(3) = 3sqrt(3) Rationalizing the denominator: () = 3sqrt(3) sqrt(3)sqrt(3) = 3sqrt(3)3 = sqrt(3)

From standard trigonometric values: (60°) = sqrt(3) Therefore: = 60°

The sun's e<=vation is 60°."}],"$L17","$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20"]

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