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QveST [7]
3 years ago
6

Aiko stands 5 m from a tree. At that distance, the angle of elevation from the ground to the top of the tree is 80°.

Mathematics
1 answer:
denis23 [38]3 years ago
3 0

Answer:

\boxed {\boxed {\sf About \ 28.36 \ meters }}

Step-by-step explanation:

Assuming the tree is perpendicular to the ground, we can use the right triangle trigonometric ratios to find the tree's height.

  • sin(θ)= opposite/hypotenuse
  • cos(θ)= adjacent/hypotenuse
  • tan(θ)= opposite/adjacent

Now, let's draw a diagram. We know Aiko is 5 meters from the base of the tree. From there, the angle to the top of the tree is 80 degrees. We are looking for x, the tree's height. The diagram attached is not to scale.

We base the sides off of the angle. x is opposite of 80 degrees and 5 is adjacent. Therefore we must use tangent.

tan(\Theta)=\frac{opposite}{adjacent}

  • opposite=x
  • adjacent=5 m
  • θ=80

Substitute in the known variables.

tan(80)=\frac{x}{5 \ m }

We want to find x, the height of the tree. Therefore we need to isolate that variable.

x is being divided and the inverse operation is multiplication. Multiply both sides of the equation by 5 meters.

5 \ m* tan(80)=\frac{x}{5 \ m }* 5 \ m

5 \ m* tan(80)=x

5 \ m *5.67128182=x

28.3564091 \ m = x

The question asks for an approximation, so let's round to the nearest hundredth.

The 6 in the thousandth place tells us to round the 5 to a 6.

28.36 \ m \approx x

The tree is about <u>28.36 meters tall.</u>

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The vertices of a triangle ABC are A(7, 5), B(4, 2), and C(9, 2). What is measure of angle ABC? 30° 45° 56.31° 78.69°
GenaCL600 [577]

The measure of angle ABC is 45°

<em><u>Explanation</u></em>

Vertices of the triangle are:   A(7, 5), B(4, 2), and C(9, 2)

According to the diagram below....

Length of the side BC (a) =\sqrt{(4-9)^2+(2-2)^2}= \sqrt{25}= 5

Length of the side AC (b) = \sqrt{(7-9)^2 +(5-2)^2}= \sqrt{4+9}=\sqrt{13}

Length of the side AB (c) = \sqrt{(7-4)^2 +(5-2)^2} =\sqrt{9+9}=\sqrt{18}

We need to find ∠ABC or ∠B . So using <u>Cosine rule</u>, we will get...

cosB= \frac{a^2+c^2-b^2}{2ac} \\ \\ cos B= \frac{(5)^2+(\sqrt{18})^2-(\sqrt{13})^2}{2*5*\sqrt{18} }\\ \\ cosB= \frac{25+18-13}{10\sqrt{18}} =\frac{30}{10\sqrt{18}}=\frac{3}{\sqrt{18}}\\ \\ cosB=\frac{3}{3\sqrt{2}} =\frac{1}{\sqrt{2}}\\ \\ B= cos^-^1(\frac{1}{\sqrt{2}})= 45 degree

So, the measure of angle ABC is 45°

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4 years ago
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