Pls. Ans. Q 26....pls. don't give link for a similar query... I want to check if my answer is correct so pls. Try Solving it

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$$\begin{gathered} Height of tall building \left(AB\right)= 20 m \\ Let, Height of the multi-storeyed building = PC \\ Angle of depression to top of building= \angle QPB= 30° \\ Angle of depression to bottom of building= \angle QPA=45° \\ Draw BD\hspace{0.17em}paralle to AC and PQ \\ Lines PQ and BD are parallel, and B\hspace{0.17em}is transversal \\ \angle PBD=\angle QPB \left(Alternate angles\right) \\ \angle PBD=30° \\ Similarly, Lines PQ and \hspace{0.17em}are parallel, and AP is the transversal \\ \angle PAC=\angle QPA \left(Alternate angles\right) \\ \angle PAC=45° \\ Now, \hspace{0.17em}and BD are parallelline \\ So, AC=BD \\ Similarly, AB and CD\hspace{0.17em}are also parallel lines. \\ So, CD=AB \\ CD =20m \\ Also, PC \hspace{0.17em}is perpendicular to AC \\ \angle PDB=\angle PCA=90° \\ In right angle △PBD \\ \tan B=\frac{PD}{BD} \\ \tan 30°=\frac{PD}{BD} \\ \frac{1}{\sqrt{3}}=\frac{PD}{BD} \\ BD=PD\sqrt{3} ....\left(1\right) \\ Similarly, In right angle △PAC \\ \tan 45°=\frac{PC}{AC} \\ AC=PC \\ BD=PC \left(as AC=BD\right) ....\left(2\right) \\ From \left(1\right) and \left(2\right) \\ PD\sqrt{3}=PC \\ PD\sqrt{3}=PD+DC \\ PD\sqrt{3}=PD+20 \\ PD\sqrt{3}-PD=20 \\ PD\left(\sqrt{3}-1\right)=20 \\ PD=\frac{20}{\left(\sqrt{3}-1\right)} \\ Rationalizing, \\ PD=\frac{20}{\left(\sqrt{3}-1\right)}\times \frac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)} \\ PD=\frac{20 \left(\sqrt{3}+1\right)}{{\left(\sqrt{3}\right)}^2-1^2} \\ PD=\frac{20 \left(\sqrt{3}+1\right)}{2} \\ PD=10 \left(\sqrt{3}+1\right) \\ Now, PC=PD+DC \\ = 10 \left(\sqrt{3}+1\right)+20 \\ = 10\sqrt{3}+30 \\ =10\left(\sqrt{3}+3\right) \left[\sqrt{3}=1.732\right] \\ =10\times 4.732 \\ =47.32 m \\ So, height of multi-storeyed building = 47.32 m \end{gathered}$$
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