ICSE Class 10 Answered

$begin mathsize 12px style straight A space vertical space pole space fixed space to space the space ground space is space divided space in space the space ratio space 1 colon 9 space by space straight a space mark space on space it space with space lower space part space shorter space than space the space upper space part. space If space the space two space parts space subtend space equal space angles space at space straight a space place space on space the space ground comma space 15 space straight m space away space from space the space base space of space the space pole comma space what space is space the space height space of space the space pole ? end style$ Please provide solution to this question

Asked by Topperlearning User | 02 Nov, 2017, 09:56: AM

Expert Answer

$begin mathsize 12px style Let space CB space be space the space pole space and space point space straight D space divides space it space such space that space BD space colon space DC space equals space 1 space colon space 9 space space Given space that space AB space equals space 15 space straight m space space Let space the space the space two space parts space subtend space equal space angles space at space point space straight A space such space that space angle CAD space equals space angle BAD space equals space straight theta space space space from space increment DAB tanθ equals BD over 15 rightwards arrow circle enclose 1 from space increment CAB tan 2 straight theta equals CB over 15 rightwards arrow circle enclose 2 on space taking space the space ratio space fraction numerator circle enclose 1 over denominator circle enclose 2 end fraction space we space get fraction numerator tanθ over denominator tan 2 straight theta end fraction equals BD over CB equals 1 over 10 rightwards arrow circle enclose 3 rightwards double arrow 10 tanθ equals fraction numerator 2 tanθ over denominator 1 minus tan squared straight theta end fraction rightwards double arrow tan squared straight theta equals 8 over 10 equals 4 over 5 rightwards double arrow tanθ equals fraction numerator 2 over denominator square root of 5 end fraction from space circle enclose 3 space we space have space tan 2 straight theta equals 10 tanθ equals fraction numerator 20 over denominator square root of 5 end fraction from space circle enclose 2 we space have space tan 2 straight theta equals CB over 15 rightwards double arrow CB equals 15 cross times fraction numerator 20 over denominator square root of 5 end fraction equals 60 square root of 5 straight m end style$