Because of prevailing winds, a tree grew so that it was leaning 6º from the vertical. At a point d = 37 meters from the tree, the angle of elevation to the top of the trees is a = 29° (see figure). Find the height h of the tree. (Round your answer to one decimal place.)
h =
m
d m
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Height of the tree is 40.0 meters.

Step-by-step explanation:

From ΔABC,

m∠ABC = 90°

Since, m∠ABC = m∠CBD + m∠ABD

90° = 6° + m∠ABD

m∠ABD = 90°- 6° = 84°

By triangle sum theorem in ΔABD,

m∠ABD + m∠BDA + m∠DAB = 180°

84° + 29° + m∠BDA = 180°

m∠BDA = 180° - 113°

             = 67°

By sine rule in ΔABD,

h =

h = 39.98

h ≈ 40.0 meters

Therefore, height of the tree is 40.0 meters.

The coordinates of point H' are (-2 , 0) ⇒ answer B

Step-by-step explanation:

* Lets revise the vertical stretch

- A vertical stretching is the stretching of the graph away from the  

 x-axis  

- If k > 1, then the graph of y = k • f(x) is the graph of f(x) vertically

 stretched by multiplying each of its y-coordinates by k

* Lets solve the problem

- Square EFGH stretches vertically by a factor of 2.5 to create

 rectangle E′F′G′H′

∴ k = 2.5

- The square stretches with respect to the x-axis

∴ The square stretches vertically

∴ The y-coordinates of each vertex of the square EFGH are

  multiplied by 2.5 to get the vertices of the rectangle E'F'G'H'

∵ Point H located at (-2 , 0)

∵ The image of point (x , y) after stretched vertically by k is (x , ky)

∴ Point H' located at (-2 , 0 × 2.5) ⇒ (-2 , 0)

∴ The coordinates of point H' are (-2 , 0)

∴ Point H' located at (-2 , 0)