Given diagram, as we can see the pole above the tower makes an angle of 58°. Since, the pole makes an angle, the trigonometric ratio we will be using is:
[tex]\boxed{\tan \theta = \frac{\text{heigh of the pole + height of the tower}}{\text{length of the ground}}}[/tex]
To find the height of the tower, first is we need to find the length of the ground. Since the problem does not give us the length of the ground, we will make a trigonometric for this.
[tex]\tan 58 = \frac{(50 + x)}{\text{length of the ground}} \\ \\ 1.6 = \frac{(50+x)}{\text{length of the ground}} \\ \\ \boxed{\text{length of the ground} = \frac{50 + x}{1.6}} \longrightarrow (1)[/tex]
Now, we will another trigonometric ratio for the angle of the tower, using the trigonometric ratio:
[tex]\boxed{\tan \theta = \frac{\text{heigh of the tower}}{\text{length of the ground}} }[/tex]
[tex]\Bigg\{\left(\begin{matrix} \tt \purple {x = height \: of \: the \: building }\\ \tt{ \purple {y = is \: the \: common \: side}}\end{matrix}\right)[/tex]
dinourawkeshi
[tex] Sige, let me check the question. [/tex]
Mikey005
thanks nag post po ako ng new may nagsagot po kasi ng hindi malinaw
dinourawkeshi
[tex] Sorry, I am still reviewing pa ng mga lesson ko since may test kami, no worries if may free time it may take 1 hour siguro since 3 question okay? [/tex]
Answers & Comments
Answer:
Hence, the height of the building is 75
Step-by-step explanation:
Given diagram, as we can see the pole above the tower makes an angle of 58°. Since, the pole makes an angle, the trigonometric ratio we will be using is:
To find the height of the tower, first is we need to find the length of the ground. Since the problem does not give us the length of the ground, we will make a trigonometric for this.
[tex]\tan 58 = \frac{(50 + x)}{\text{length of the ground}} \\ \\ 1.6 = \frac{(50+x)}{\text{length of the ground}} \\ \\ \boxed{\text{length of the ground} = \frac{50 + x}{1.6}} \longrightarrow (1)[/tex]
Now, we will another trigonometric ratio for the angle of the tower, using the trigonometric ratio:
We have,
[tex]\tan 44 = \frac{x}{\text{length of the ground}} \\ \\ 0.96 = \frac{x}{L} \\ \\ 0.96L=h \\ 96L=h\cdot \:100\\ \\ L=\frac{25h}{24} \longrightarrow (2)[/tex]
(note: L stands for the length of the ground)
Now that we found the length of the ground we can now solve for the height of the tower.
[tex](1) = (2) \\ \\ \frac{50+x}{1.6}=\frac{25x}{24} \\ \\ \left(50+x\right) \:24=1.6\cdot \:25x \\ \\ \left(50+x\right) \:24=40x \\ \\ 1200+24x=40x \\ \\ -16x=-1200 \\ \\ x=75[/tex]
Hence, the height of the building is 75
Verified answer
HEIGHT OF THE BUILDING
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[tex]\begin{gathered}{\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} \\\end{gathered}[/tex]
[tex]\begin{gathered} \qquad \qquad \quad\qquad \large \sf{ \bold{ 76.09 \: \: feet}} \\ \end{gathered} [/tex]
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[tex]\begin{gathered}{\underline{\huge \mathbb{S} {\large \mathrm {OLUTION : }}}} \\\end{gathered} [/tex]
[tex] \bullet \: \tt{ Using \: the \: tangent \: trigonometric \: function \: \purple{TOA }}[/tex]
[tex]\tt{from \: \purple{ SO \: CAH \: TOA }}[/tex]
[tex]\begin{gathered}\qquad\qquad\sf\hookrightarrow \: \boxed{\bold{ \: \: \: \tan \theta =\dfrac{opposite}{adjacent} \: \: \: }} \\ \end{gathered} [/tex]
[tex]\begin{gathered} \qquad \quad\sf\: \tan 44^{\circ}= \dfrac{opposite}{adjacent} \\ \end{gathered} [/tex]
[tex]\tt{When \: \theta = \purple{44^{\circ}:}}[/tex]
[tex]\begin{gathered} \qquad \quad\sf \: \tan 44 {}^{ \circ} = \bold{ \dfrac{x}{y}} \\ \end{gathered} [/tex]
[tex]\begin{gathered} \qquad \quad\sf\implies \: y = \dfrac{x }{\tan 44^{\circ}} \longrightarrow \bold{I} \\ \end{gathered} [/tex]
[tex]\Bigg\{\left(\begin{matrix} \tt \purple {x = height \: of \: the \: building }\\ \tt{ \purple {y = is \: the \: common \: side}}\end{matrix}\right)[/tex]
[tex]\qquad \qquad{ \overline{ \qquad \qquad \qquad \qquad \qquad}} [/tex]
[tex]\tt{When \: \theta = \purple{58^{\circ}:}}[/tex]
[tex]\begin{gathered} \qquad \quad\sf\implies \: \tan 58^{\circ} = \dfrac{50 + x}{y} \\ \end{gathered} [/tex]
[tex]\begin{gathered} \qquad \quad\sf\implies \: y = \dfrac{x }{\tan 44^{\circ}} \longrightarrow \bold{II} \\ \end{gathered} [/tex]
[tex]\tt{Now, \: equate \: \purple{(I)} \: and \purple{(II)} \: to \: solve \: for \: x \: as \: follows:}[/tex]
[tex]\begin{gathered} \qquad \quad\sf \: \dfrac{x}{ \tan {44}^{ \circ} } = \dfrac{50 + x}{ \tan {58}^{ \circ} } \\ \end{gathered} [/tex]
[tex]\begin{gathered} \qquad \quad\sf\: \dfrac{x}{ 0.9657} = \dfrac{50 + x}{ 1.6003} \\ \end{gathered} [/tex]
[tex]\tt{Cross \: multiply \: : }[/tex]
[tex]\begin{gathered} \qquad \quad\sf \: 1.6003x = 0.9657(50 + x) \\ \end{gathered} [/tex]
[tex]\begin{gathered} \qquad \quad\sf\: 1.6003x = \bold{48.285 + 0.9657x} \\ \end{gathered} [/tex]
[tex]\tt{Combine \: like \: terms: }[/tex]
[tex]\begin{gathered} \qquad \quad\sf\: 1.6003 - 0.9657x = 48.285 \\ \end{gathered} [/tex]
[tex]\begin{gathered} \qquad \quad\sf\: 0.6346x = 48.285\\ \end{gathered} [/tex]
[tex]\begin{gathered} \qquad \quad\sf\: \frac{48.285}{0.6346} = 76.09\\ \end{gathered} [/tex]
[tex]\begin{gathered} \qquad \quad\sf\: \large{\boxed{\bold{x= 76.09 \: ft}}} \\ \end{gathered} [/tex]
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