The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30°. If the jet plane is flying at a constant height, then its height is:

Option 1 - <p>3600 &lt;!-- [if gte mso 9]>&lt;xml&gt; <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1816608243"> </o:OLEObject> &lt;/xml&gt;&lt;![endif]--&gt;&nbsp;<span class="mathml" contenteditable="false"> <math> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mtext> </mtext> <mi>m</mi> </mrow> </math> </span></p>
Option 2 - <p>&lt;!-- [if gte mso 9]>&lt;xml&gt; <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1816608251"> </o:OLEObject> &lt;/xml&gt;&lt;![endif]--&gt;&nbsp;<span class="mathml" contenteditable="false"> <math> <mrow> <mn>2</mn> <mn>4</mn> <mn>0</mn> <mn>0</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mtext> </mtext> <mi>m</mi> </mrow> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>1</mn> <mn>2</mn> <mn>0</mn> <mn>0</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mtext> </mtext> <mi>m</mi> </mrow> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>1</mn> <mn>8</mn> <mn>0</mn> <mn>0</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mtext> </mtext> <mi>m</mi> </mrow> </math> </span></p>
4 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
A
7 months ago
Correct Option - 3
Detailed Solution:

S = 432 km/hr

tan 60° =   h y

h = 3 y

t a n 3 0 ° = h x + y           

? x = 4 3 2 * 1 0 0 0 3 6 0 0 * 2 0           

x = 2400 m

y = 1200 m

h = 1 2 0 0 3 m          

 

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Maths Ncert Solutions class 11th 2026

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