A man 1.7m tall observes a bird on top of a tree at an angle of 30°. if the distance between the man's head and the bird is 25m, what is the height of the tree?
26.7m
14.2m
\(1.7+(25\frac{\sqrt{3}}{2}m\)
\(1.7+(25\frac{\sqrt{2}}{2}m\)
Explanation
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Discussions (21)
yes b is the answer / |
heres the diagram / |
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25 / | h
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/30 |
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1.7 | | 1.7
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the reason why here is 1.7 on the both sides is because it is a rectangle
and u know on a rectangle on two of the sides are equal
so therefore to find h using SOHCATOA
we use sine which is opposite over hypotenuse
so that is sin30=h/25
=25xsin30=12.5
but thats not all
we have only found h not H so to find H all u have to do is add 12.5 to 1.7 and that will give u 14.2
B
glad i helped
if there is any problem with my solving kindly comment
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25m x |
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___30¤_______________________________ |
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|1.7m
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using sine rule
25/sin90¤ = x/sin30¤
where: sin90¤=1 and sin30¤=1/2
therefore
25/1=x/ 1/2
25/1=x/1 / 1/2
25/1=x/1 * 2/1
25=2x
divide both sides by 2
x=13.5m
Therefore the height of the three =1.7m 13.5m
=14.2m
the answer is B
height of man = 1.7
head to bird distance = 25
sin 30° = 1/2
head level to tree top = x
SOHCAHTOA
using SOH
Sine 30° = x/25
1/2=x/25
x = 12.5
height of tree = head level to tree top (x) ground level to head Level (height of man = 1.7)
= 12.5 1.7 = 14.2m
myschool.com.ng = wrong
kindly change the answer, some might be misled
The connection hia is bw hyp n opp which gives SOH Ie sin &( nt opp n adj as in TOA Ie tan) bc d distance bw d bird n d man's head is d hyp.
25 /|x
1.7 | |1.7
D4, D height is x 1.7
sin 30=x/25.
sin 30 is 1/2.
Ie 1/2 =x/25
crossmutiplyin, we v
2x =25
x =25/2 =12.5
so x 1.7 is nw
12.5 1.7 = 14.2
yep i go wit wheezboi nd dose dt get 14.2m bcos after getting 12.5m u still need 2 b add 2 1.7m inorder 2 get whole height of d tree
if you do your diagram well
you realise you were given your opposite and adjacent
therefore the trig is tan
tan30=x/25
x=25/√(3)
which become
x=25√(3)/3
height of tree = 1.7 + 25√(3)/3
Firstly sin 30°= o.5 just multiply by the distance of the man 25m then you add the height of the man so: 0.5×25 =12.5, 12.5+ 1.7 = 14.2
That's what I understand
Distance btw d man's head and THE BIRD (nt d tree) is 25m... This should be d hypotenus side nd nt d adjacent side. With this, you'll arrive at 14.2m instead. The sketch could be wrong then
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| \ let here be x| \
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|-------------30\|
1.7 |..................................| d distance is 25m using pathg theory x is d opp bcos dat is where are angle Θ is facin and distance is ajc usin SOHCAHTOA tanΘ is =opp/ajc by spceial triangle ratio tan30 is root 3/3=opp which is x/ajc which is 25 we now hav root 3/3=x/25 cross multiply we hav 25*root 3=3x i.e x is =25*rot 3/3 plus d height of d man which is 1.7 ur answer wil now be 1.7 25*root 3/3
height of d tree from d man's head 2 d top 's 25m,nd d man height is 1.7 then d height of d tree must be 25+1.7=26.7.
