A. Find the height of the mountain; B. Find the distance to the nearest kilometre stone
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From the top of a mountain, in the same vertical plane is two consecutive kilometers stones on the level piece of ground the angles of depression to the kilometer stones are 42ð 12 minutes and 23ð 30 minutes respectively.
A. Calculate the hate of the mountain
B. Calculate the distance to the nearest kilometre stone
trigonometry
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From the top of a mountain, in the same vertical plane is two consecutive kilometers stones on the level piece of ground the angles of depression to the kilometer stones are 42ð 12 minutes and 23ð 30 minutes respectively.
A. Calculate the hate of the mountain
B. Calculate the distance to the nearest kilometre stone
trigonometry
"height" not hate.
â David Diaz
Jul 20 at 16:02
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
From the top of a mountain, in the same vertical plane is two consecutive kilometers stones on the level piece of ground the angles of depression to the kilometer stones are 42ð 12 minutes and 23ð 30 minutes respectively.
A. Calculate the hate of the mountain
B. Calculate the distance to the nearest kilometre stone
trigonometry
From the top of a mountain, in the same vertical plane is two consecutive kilometers stones on the level piece of ground the angles of depression to the kilometer stones are 42ð 12 minutes and 23ð 30 minutes respectively.
A. Calculate the hate of the mountain
B. Calculate the distance to the nearest kilometre stone
trigonometry
asked Jul 20 at 15:58
Naftali PNA Shikongo
11
11
"height" not hate.
â David Diaz
Jul 20 at 16:02
add a comment |Â
"height" not hate.
â David Diaz
Jul 20 at 16:02
"height" not hate.
â David Diaz
Jul 20 at 16:02
"height" not hate.
â David Diaz
Jul 20 at 16:02
add a comment |Â
1 Answer
1
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Denote $alpha = 42^circ12'$, $beta = 23^circ30'$.
$h - $ height; $d - $ distance to nearest km stone.
Then
$$tan alpha = dfrachd; tag1$$
$$tan beta = dfrachd+1. tag2$$
So :
$$
dfractanalphatanbeta = dfrach/dh/(d+1)=dfracd+1d =1+dfrac1d.
$$
And therefore
$$
d = dfrac1dfractanalphatanbeta - 1=dfractanbetatanalpha-tanbeta.tag3
$$
Then $h$ can be found easily from $(1), (3)$.
1
Thank you very much Oleg
â Naftali PNA Shikongo
Jul 21 at 16:51
@NaftaliPNAShikongo If Oleg's answer was helpful, please upvote or accept it by clicking the up arrow or the checkmark, respectively, near the upper-left corner of the answer.
â Robert Howard
Aug 10 at 19:48
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Denote $alpha = 42^circ12'$, $beta = 23^circ30'$.
$h - $ height; $d - $ distance to nearest km stone.
Then
$$tan alpha = dfrachd; tag1$$
$$tan beta = dfrachd+1. tag2$$
So :
$$
dfractanalphatanbeta = dfrach/dh/(d+1)=dfracd+1d =1+dfrac1d.
$$
And therefore
$$
d = dfrac1dfractanalphatanbeta - 1=dfractanbetatanalpha-tanbeta.tag3
$$
Then $h$ can be found easily from $(1), (3)$.
1
Thank you very much Oleg
â Naftali PNA Shikongo
Jul 21 at 16:51
@NaftaliPNAShikongo If Oleg's answer was helpful, please upvote or accept it by clicking the up arrow or the checkmark, respectively, near the upper-left corner of the answer.
â Robert Howard
Aug 10 at 19:48
add a comment |Â
up vote
0
down vote
Denote $alpha = 42^circ12'$, $beta = 23^circ30'$.
$h - $ height; $d - $ distance to nearest km stone.
Then
$$tan alpha = dfrachd; tag1$$
$$tan beta = dfrachd+1. tag2$$
So :
$$
dfractanalphatanbeta = dfrach/dh/(d+1)=dfracd+1d =1+dfrac1d.
$$
And therefore
$$
d = dfrac1dfractanalphatanbeta - 1=dfractanbetatanalpha-tanbeta.tag3
$$
Then $h$ can be found easily from $(1), (3)$.
1
Thank you very much Oleg
â Naftali PNA Shikongo
Jul 21 at 16:51
@NaftaliPNAShikongo If Oleg's answer was helpful, please upvote or accept it by clicking the up arrow or the checkmark, respectively, near the upper-left corner of the answer.
â Robert Howard
Aug 10 at 19:48
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Denote $alpha = 42^circ12'$, $beta = 23^circ30'$.
$h - $ height; $d - $ distance to nearest km stone.
Then
$$tan alpha = dfrachd; tag1$$
$$tan beta = dfrachd+1. tag2$$
So :
$$
dfractanalphatanbeta = dfrach/dh/(d+1)=dfracd+1d =1+dfrac1d.
$$
And therefore
$$
d = dfrac1dfractanalphatanbeta - 1=dfractanbetatanalpha-tanbeta.tag3
$$
Then $h$ can be found easily from $(1), (3)$.
Denote $alpha = 42^circ12'$, $beta = 23^circ30'$.
$h - $ height; $d - $ distance to nearest km stone.
Then
$$tan alpha = dfrachd; tag1$$
$$tan beta = dfrachd+1. tag2$$
So :
$$
dfractanalphatanbeta = dfrach/dh/(d+1)=dfracd+1d =1+dfrac1d.
$$
And therefore
$$
d = dfrac1dfractanalphatanbeta - 1=dfractanbetatanalpha-tanbeta.tag3
$$
Then $h$ can be found easily from $(1), (3)$.
answered Jul 20 at 16:08
Oleg567
13.6k22969
13.6k22969
1
Thank you very much Oleg
â Naftali PNA Shikongo
Jul 21 at 16:51
@NaftaliPNAShikongo If Oleg's answer was helpful, please upvote or accept it by clicking the up arrow or the checkmark, respectively, near the upper-left corner of the answer.
â Robert Howard
Aug 10 at 19:48
add a comment |Â
1
Thank you very much Oleg
â Naftali PNA Shikongo
Jul 21 at 16:51
@NaftaliPNAShikongo If Oleg's answer was helpful, please upvote or accept it by clicking the up arrow or the checkmark, respectively, near the upper-left corner of the answer.
â Robert Howard
Aug 10 at 19:48
1
1
Thank you very much Oleg
â Naftali PNA Shikongo
Jul 21 at 16:51
Thank you very much Oleg
â Naftali PNA Shikongo
Jul 21 at 16:51
@NaftaliPNAShikongo If Oleg's answer was helpful, please upvote or accept it by clicking the up arrow or the checkmark, respectively, near the upper-left corner of the answer.
â Robert Howard
Aug 10 at 19:48
@NaftaliPNAShikongo If Oleg's answer was helpful, please upvote or accept it by clicking the up arrow or the checkmark, respectively, near the upper-left corner of the answer.
â Robert Howard
Aug 10 at 19:48
add a comment |Â
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"height" not hate.
â David Diaz
Jul 20 at 16:02