If the angles of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with it are complementary, then the height of the tower is

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KollegeApply · Accepted Answer

Let h be the height of tower AB. Given that: angle of elevation of top of the tower are $\angle$D=$\theta$and .$\angle$C=$90^\circ$-$\theta$ DistanceBC=b and BD=a Here, we have to find the height of tower. So we use trigonometric ratios. In a triangle, ABC tan D=(AB)/(BC) $\Rightarrow$ tan ($90^\circ$-$\theta$)=h/b $\Rightarrow$ cot$\theta$=h/b Again in a triangle ABD, tan D=(AB)/(BD) $\Rightarrow$ tan $\theta$=h/a $\Rightarrow$1/cot $\theta$=h/a $\Rightarrow$ b/h=h/a $\Rightarrow$h^2=ab $\Rightarrow$ h=sqrt(ab) Put cot$\theta$=h/b [Solution diagram: 63492_solution_1.png]

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