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Answered Same Day Dec 23, 2021

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David answered on Dec 23 2021

131 Votes

Problem 1
Solution:
Let height of tree from mr thomas’s eyes level is ‘h’ and 20 feet is distance between mr Thomas and
tree. In the right angle triangle in the above figure,
tan(67°)= h/20
 1.75=h/20 (tan(67°)=1.75)
 h=35 feet
total height of tree= 35+5.67=40.67
tan(q)=(40.67-5)/10=3.565
Angle of elevation , q=82.58°
Su
ounding that is susceptible to trees fall is circle with radius equals height of tree
Area of su
ounding=π*40.67*40.67=5196.3 feet *feet
It’s a huge area and tree may fall in any direction. The chances of falling tree in the particular
direction are very low. Its not very reasonable request.
67
20 feet
68 inches=
68/12=
5.67 feet
h
From law of cosines
L2=642+1022-2*64*102*cos(52°)
L2=5562.5
L=74.58 yards
One of the way is similar to the above one . Thomas can measure distance from two endpoints of width
to a common point and find angle between these lines and apply cosine law.
Problem 2 The race
1 )
Ans let the distance by race by sea course is d
Since there is right angles triangle formed. By Pythagoras theorem , this d2= sqrt(2.5*2.5-2*2)
64 yards
102 yards
L=length of the
lake
 d=1.5 mi
Now, who will win depend on the speed of racers on sea and land respectively. Obviously the sea
acer can afford to go slower than land racer.
2)
Ans let the distance by race by sea course is d
Since there is right angles triangle formed. By Pythagoras theorem , this d2= sqrt(2.5*2.5-2*2)
 d=1.5...

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