Question

From a window 20m above the ground, a person could see the top of the house across the street at an angle of elevation of 5 degrees. The angle of depression to the base of the house was 15 degrees. What is the height of the house?

Answer 1

Answer:

To find:  Height of the house

Tan 15 ° = 0.2679

Tan 5 ° = 0.0875

BC=AD

Step-by-step explanation:

In triangle ABC,                  

Tan 15 ° = AB/BC

0.2679 = 20/BC

BC = 20/0.2679

BC = 74.65 m

AC² =  AB²+ BC² ( Pythagoras theorem )

AC² = 20² + 74.65²

AC² = 400 + 5572.6

AC² = 5972.6

In triangle ADE,

Tan 5 ° = DE/AD

0.0875 = DE/BC    ( AD = BC )

BC * 0.0875 = DE

74.65 * 0.0875 = DE

DE = 6.53 m

DC² = AC² - AD²

DC² = 5972.6 - 5572.6

DC² = 400

DC = 20 m

So therefore the height of the house is ED + DC

ED + DC = 6.53 m + 20 m

EC = 26.53 m ( ED + DC = EC = HEIGHT OF THE HOUSE )

I hope you understood the answer

Answer 2

Answer:

26.53 m

Step-by-step explanation:

Given:

A person could view the top of the home across the street at an elevation of 5 degrees from a window 20 metres above the ground. The house's base was 15 degrees away from the depression.

To find:

Height

Solution:

An essential element in trigonometric calculations is the angle of depression. When the observer is positioned higher than the item they are attempting to examine, an angle of depression is created. A line connecting the object and the observer's eyes (the line of sight) can be drawn when the observer is looking down at the item. A second horizontal line can be drawn starting at the observer's eye level. The term "angle of depression" refers to the angle created between the horizontal line and the line of sight. The angle of elevation and depression is the same.

From the given figure,

In ΔABC

$tan 15^{o} =\frac{AB}{BC} \\0.2679=\frac{20}{BC} \\BC=\frac{20}{0.2679} \\BC=74.65m$

By using  Pythagoras theorem

$AC^{2} =AB^{2} +BC^{2} \\AC^{2}=20^{2}+74.65^{2}\\AC^{2}=400+5572.6\\AC^{2}=5972.6$

Now in ΔADE

$Tan5^{o} = \frac{DE}{AD} \\0.0875=\frac{DE}{AD} (AD=BC)\\\\BC(0.0875)=DE\\(74.65)(0.0875)=DE\\DE=6.53m$

$DC^{2}= AC^{2} -AD^{2} \\DC^{2}= 5972.6 - 5572.6\\DC^{2}=400\\DC=20m$

Thus the height is ED+DC

$ED + DC = 6.53 m + 20 mEC = 26.53 m$

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