Q. 44.0( 4 Votes )

In the picture, the vertical lines are equally spaced. Prove that their heights are in arithmetic sequence. What is the common difference?


Answer :

Let us label the diagram



Let the distance between two consecutive vertical lines be 'x' and height of vertical lines be h1, h2, h3, …, and so on.


Let AB1 = x


Now, we know



perpendicular = base × tanθ


Therefore, in consecutive triangles, we have


B1C1 = AB1 × tan 40°


h1 = atan40°


B2C2 = AB2 × tan 40°


h2 = (a + x)tan40°


B3C3 = AB3 × tan 40°


h3 = (a + 2x)tan40°


.


.


.


and so on.


Now, we have


h1, h2, h3,… = atan40°, (a + x)tan40°, (a + x)tan40°, …


Let us calculate the common difference.


h2 - h1 = (a + x)tan40° - atan40° = xtan40°


h3 - h2 = (a + 2x)tan40° - (a + x)tan40° = xtan40°


As, the common difference between the terms is same, terms are in AP and hence,


h1, h2, h3, …. are in AP.


Therefore, heights of vertical lines are in AP with common difference xtan40°, where 'x' is the space between two consecutive lines. [tan 40° = 0.8391]


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