Answer:
[tex] \qquad \large \bold{Distance \: from \: B = \pmb{19.5 \: m}, } \\ [/tex]
[tex] \qquad \large \bold{Height \: of \: clock \: tower = \pmb{ 9.1 \: m}} \\ [/tex]
Step-by-step explanation:
The given scenario can be modeled as two right triangles that have the same height (see attachment).
[tex] \\ \underline{ \sf{ \pmb{Define \: the \: variables : }}} \\ [/tex]
- Let x = the distance between point B and the foot of the clock tower.
- Let y = the height of the clock tower.
Use the given information and the tan trigonometric ratio to create equations for the height of the clock tower.
[tex] \\ \underline{ \sf{ \pmb{Tan \: trigonometric \: ratio : }}} \\ [/tex]
[tex] \qquad \qquad \leadsto \: {\sf \tan(\theta)=\dfrac{O}{A}} \\ [/tex]
[tex] \sf{where:} \\ [/tex]
- [tex] \sf{ \pmb{\theta }\: is \: the \: \pmb{angle}} \\ [/tex]
- [tex] \sf{ \pmb{O} \: is \: the \: \pmb{ side \: opposite} \: the \: \pmb{ angle}} \\ [/tex]
- [tex] \sf{ \pmb{A} \: is \: the \: \pmb{side \: adjacent} \: of\: the \: \pmb{ angle}} \\ [/tex]
[tex] \underline{ \sf{ \pmb{For \: triangle \: (point \: A):}}} \\ [/tex]
- [tex] \bold{ \theta = 22°} \\ [/tex]
- [tex] \bold{O = y} \\ [/tex]
- [tex] \bold{A = 3 + x} \\ [/tex]
[tex] \large \sf{Therefore:} \\ [/tex]
[tex]\implies \: \sf \tan(22^{\circ})=\dfrac{y}{3+x} \\ [/tex]
[tex]\implies \: \sf y=(3+x)\tan(22^{\circ}) \\ [/tex]
[tex] \underline{ \sf{ \pmb{For \: triangle \: (point \: B):}}} \\ [/tex]
- [tex] \bold{ \theta = 25°} \\ [/tex]
- [tex] \bold{O = y} \\ [/tex]
- [tex] \bold{A = x} \\ [/tex]
[tex] \large \sf{Therefore:} \\ [/tex]
[tex]\implies \: \sf \tan(25^{\circ})=\dfrac{y}{x} \\ [/tex]
[tex]\implies \: \sf y=x\tan(25^{\circ}) \\ [/tex]
Substitute the equation for Triangle A into the equation for Triangle B and solve for x:
[tex]\implies \: \sf (3+x)\tan(22^{\circ})=x\tan(25^{\circ}) \\ [/tex]
[tex]\implies \: \sf 3\tan(22^{\circ})+x\tan(22^{\circ})=x\tan(25^{\circ}) \\ [/tex]
[tex]\implies \: \sf 3\tan(22^{\circ})=x\tan(25^{\circ})-x\tan(22^{\circ}) \\ [/tex]
[tex]\implies \: \sf 3\tan(22^{\circ})=x[\tan(25^{\circ})-\tan(22^{\circ}) \\ [/tex]
[tex]\implies \: \sf x=\dfrac{3\tan(22^{\circ})}{\tan(25^{\circ})-\tan(22^{\circ})} \\ [/tex]
[tex]\implies \: \bold{ x= \pmb{19.46131668. . .m}} \\ \\[/tex]
Therefore, the distance of B to the foot of the clock tower is 19.5 m (nearest tenth).
To find the height of the clock tower, substitute the found value of x into one of the found equations for y:
[tex]\implies \: \sf y=x\tan(25^{\circ}) \\ [/tex]
[tex]\implies \: \sf y=\left(\dfrac{3\tan(22^{\circ})}{\tan(25^{\circ})-\tan(22^{\circ})}\right)\tan(25^{\circ}) \\ [/tex]
[tex]\implies \: \sf y=\dfrac{3\tan(22^{\circ})\tan(25^{\circ})}{\tan(25^{\circ})-\tan(22^{\circ})} \\ [/tex]
[tex]\implies \: \bold{ y= \pmb{9.074961005. . .m}} \\ \\ [/tex]
Therefore, the height of the clock tower is 9.1 m (nearest tenth).
