Answer:
14.5 m
Step-by-step explanation:
Here it is given that a tower stands on level ground and from a point P , the angle of elevation is 26° . And there's another point Q, 3m above point P from where the angle of elevation is 21° . We need to find out the height of the tower.
Figure:
[tex] \begin{picture}(8,8)\setlength{\unitlength}{1cm}\linethickness{.4mm}\put(0,0){\line(1,0){4}}\put(4,0){\line(0,1){3}}\put(0,0.01){\line(4,3){4}}\put(4,3.01){\line(-2, - 1){4}}\put(0,0.01){\line(0,1){1}}\put(0,1){\line(1,0){4}}\put(4.3,2){$\sf x $}\put(4.3,0.5){$\sf 3m $}\put(0, - 0.4){$\sf P \: $}\put(4, - 0.5){$\sf \: B $}\put(4.3,2){$\sf x $}\put(4,3.2){$\sf A $}\put(4.3,2){$\sf x $}\put(1,1.2){$\sf 21^\circ $}\put(0,1.2){$\sf Q $}\put(1,.2){$\sf 26^\circ $}\put(4,4){$\boxed{\sf - RISH4BH }$ }\put(2,.7){$\sf y $}\put(4.4,1){$\sf C $}\put(2, - .2){$\sf y $}\end{picture}[/tex]
So , in ∆ABP , we have ;
[tex]\longrightarrow tan26^o =\dfrac{AB}{BP}[/tex]
Substitute value of tan26° = 0.48 ,
[tex]\longrightarrow 0.48 =\dfrac{x+3}{y}\\[/tex]
[tex]\longrightarrow 0.48y = x + 3 \dots . (i)[/tex]
Again, in ∆AQC , we have ;
[tex]\longrightarrow tan21^o = \dfrac{AC}{QC} [/tex]
Substitute the value of tan21°= 0.38 ,
[tex]\longrightarrow 0.38 =\dfrac{x}{y}\\[/tex]
[tex]\longrightarrow x = 0.38y \dots .(ij)[/tex]
From equation (i) and (ii) , we have ;
[tex]\longrightarrow 0.48 \times \dfrac{x}{0.38}= x +3 \\ [/tex]
[tex]\longrightarrow 1.26x = x +3\\ [/tex]
[tex]\longrightarrow 1.26x - x = 3\\ [/tex]
[tex]\longrightarrow 0.26x = 3 \\ [/tex]
[tex]\longrightarrow x = \dfrac{3}{0.26}\\ [/tex]
[tex]\longrightarrow \underline{\underline{ x = 11.5\ m }} [/tex]
Therefore the height of the tower will be ,
[tex]\longrightarrow height = x + 3m \\ [/tex]
[tex]\longrightarrow \underline{\underline{ Height = 14.5 \ m }} [/tex]
