Mixing
math question angle of elevation
$begingroup$
A tree is $x$ meters high. The angle of elevation of its top from a point $P$ on the ground is 23 degrees. From another point $Q$, 10 meters from $P$ and in line with $P$ and the foot of the tree, the angle of elevation is 32 degrees. Find $x$.
[OP points out in the comments: if $QR=y$, then $tan32=x/y$, so $y=x/tan32$, and $tan23=x/(10+y)$.]
trigonometry
edited Nov 17 '14 at 2:33
Gerry Myerson
147k8151304
asked Nov 16 '14 at 8:51
Asad RafiqAsad Rafiq
739
$endgroup$
add a comment |
$begingroup$
A tree is $x$ meters high. The angle of elevation of its top from a point $P$ on the ground is 23 degrees. From another point $Q$, 10 meters from $P$ and in line with $P$ and the foot of the tree, the angle of elevation is 32 degrees. Find $x$.
[OP points out in the comments: if $QR=y$, then $tan32=x/y$, so $y=x/tan32$, and $tan23=x/(10+y)$.]
trigonometry
edited Nov 17 '14 at 2:33
Gerry Myerson
147k8151304
asked Nov 16 '14 at 8:51
Asad RafiqAsad Rafiq
739
$endgroup$
$begingroup$
Let QR = y. Can you use the two right angled triangles? Use them to set up two equations and solve them for y and x.
$endgroup$
– Mick
Nov 16 '14 at 10:05
$begingroup$
I am using tan32 = x/y +> y = x/tan32 and for x: tan 23 = x/10+y
$endgroup$
– Asad Rafiq
Nov 16 '14 at 10:12
add a comment |
$begingroup$
A tree is $x$ meters high. The angle of elevation of its top from a point $P$ on the ground is 23 degrees. From another point $Q$, 10 meters from $P$ and in line with $P$ and the foot of the tree, the angle of elevation is 32 degrees. Find $x$.
[OP points out in the comments: if $QR=y$, then $tan32=x/y$, so $y=x/tan32$, and $tan23=x/(10+y)$.]
trigonometry
edited Nov 17 '14 at 2:33
Gerry Myerson
147k8151304
asked Nov 16 '14 at 8:51
Asad RafiqAsad Rafiq
739
$endgroup$
A tree is $x$ meters high. The angle of elevation of its top from a point $P$ on the ground is 23 degrees. From another point $Q$, 10 meters from $P$ and in line with $P$ and the foot of the tree, the angle of elevation is 32 degrees. Find $x$.
[OP points out in the comments: if $QR=y$, then $tan32=x/y$, so $y=x/tan32$, and $tan23=x/(10+y)$.]
trigonometry
trigonometry
edited Nov 17 '14 at 2:33
Gerry Myerson
147k8151304
asked Nov 16 '14 at 8:51
Asad RafiqAsad Rafiq
739
edited Nov 17 '14 at 2:33
Gerry Myerson
147k8151304
asked Nov 16 '14 at 8:51
Asad RafiqAsad Rafiq
739
edited Nov 17 '14 at 2:33
Gerry Myerson
147k8151304
edited Nov 17 '14 at 2:33
Gerry Myerson
147k8151304
edited Nov 17 '14 at 2:33
Gerry Myerson
147k8151304
147k8151304
asked Nov 16 '14 at 8:51
Asad RafiqAsad Rafiq
739
asked Nov 16 '14 at 8:51
Asad RafiqAsad Rafiq
739
asked Nov 16 '14 at 8:51
Asad RafiqAsad Rafiq
739
739
$begingroup$
Let QR = y. Can you use the two right angled triangles? Use them to set up two equations and solve them for y and x.
$endgroup$
– Mick
Nov 16 '14 at 10:05
$begingroup$
I am using tan32 = x/y +> y = x/tan32 and for x: tan 23 = x/10+y
$endgroup$
– Asad Rafiq
Nov 16 '14 at 10:12
add a comment |
$begingroup$
Let QR = y. Can you use the two right angled triangles? Use them to set up two equations and solve them for y and x.
$endgroup$
– Mick
Nov 16 '14 at 10:05
$begingroup$
I am using tan32 = x/y +> y = x/tan32 and for x: tan 23 = x/10+y
$endgroup$
– Asad Rafiq
Nov 16 '14 at 10:12
$begingroup$
Let QR = y. Can you use the two right angled triangles? Use them to set up two equations and solve them for y and x.
$endgroup$
– Mick
Nov 16 '14 at 10:05
$begingroup$
Let QR = y. Can you use the two right angled triangles? Use them to set up two equations and solve them for y and x.
$endgroup$
– Mick
Nov 16 '14 at 10:05
$begingroup$
I am using tan32 = x/y +> y = x/tan32 and for x: tan 23 = x/10+y
$endgroup$
– Asad Rafiq
Nov 16 '14 at 10:12
$begingroup$
I am using tan32 = x/y +> y = x/tan32 and for x: tan 23 = x/10+y
$endgroup$
– Asad Rafiq
Nov 16 '14 at 10:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let the length $PR$ be $y$. Then we have the following two equations:
In $triangle PRS$, we have $tan(23^circ)=frac{x}{y}$ $(1)$ and $tan(32^circ)=frac{x}{y-10}$ $(2)$.
Solving for $y$ in equation $2$, we get $y=frac{x}{tan(32^circ)}+10$. Plugging this into equation $1$ and solving gives $xapprox13.236$.
answered Nov 16 '14 at 10:34
rae306rae306
5,52831135
$endgroup$
add a comment |
$begingroup$
It is easy to get
begin{equation}
10 = frac{x}{tan 23^{o}}-frac{x}{tan 32^{o}}
end{equation}
and then you can obtain that
begin{equation}
x = frac{10}{frac{1}{tan 23^{o}} - frac{1}{tan 32^{o}}}
end{equation}
answered Nov 17 '14 at 3:51
askuyueaskuyue
16510
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1023958%2fmath-question-angle-of-elevation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let the length $PR$ be $y$. Then we have the following two equations:
In $triangle PRS$, we have $tan(23^circ)=frac{x}{y}$ $(1)$ and $tan(32^circ)=frac{x}{y-10}$ $(2)$.
Solving for $y$ in equation $2$, we get $y=frac{x}{tan(32^circ)}+10$. Plugging this into equation $1$ and solving gives $xapprox13.236$.
answered Nov 16 '14 at 10:34
rae306rae306
5,52831135
$endgroup$
add a comment |
$begingroup$
Let the length $PR$ be $y$. Then we have the following two equations:
In $triangle PRS$, we have $tan(23^circ)=frac{x}{y}$ $(1)$ and $tan(32^circ)=frac{x}{y-10}$ $(2)$.
Solving for $y$ in equation $2$, we get $y=frac{x}{tan(32^circ)}+10$. Plugging this into equation $1$ and solving gives $xapprox13.236$.
answered Nov 16 '14 at 10:34
rae306rae306
5,52831135
$endgroup$
add a comment |
$begingroup$
Let the length $PR$ be $y$. Then we have the following two equations:
In $triangle PRS$, we have $tan(23^circ)=frac{x}{y}$ $(1)$ and $tan(32^circ)=frac{x}{y-10}$ $(2)$.
Solving for $y$ in equation $2$, we get $y=frac{x}{tan(32^circ)}+10$. Plugging this into equation $1$ and solving gives $xapprox13.236$.
answered Nov 16 '14 at 10:34
rae306rae306
5,52831135
$endgroup$
Let the length $PR$ be $y$. Then we have the following two equations:
In $triangle PRS$, we have $tan(23^circ)=frac{x}{y}$ $(1)$ and $tan(32^circ)=frac{x}{y-10}$ $(2)$.
Solving for $y$ in equation $2$, we get $y=frac{x}{tan(32^circ)}+10$. Plugging this into equation $1$ and solving gives $xapprox13.236$.
answered Nov 16 '14 at 10:34
rae306rae306
5,52831135
answered Nov 16 '14 at 10:34
rae306rae306
5,52831135
answered Nov 16 '14 at 10:34
rae306rae306
5,52831135
answered Nov 16 '14 at 10:34
rae306rae306
5,52831135
5,52831135
add a comment |
add a comment |
$begingroup$
It is easy to get
begin{equation}
10 = frac{x}{tan 23^{o}}-frac{x}{tan 32^{o}}
end{equation}
and then you can obtain that
begin{equation}
x = frac{10}{frac{1}{tan 23^{o}} - frac{1}{tan 32^{o}}}
end{equation}
answered Nov 17 '14 at 3:51
askuyueaskuyue
16510
$endgroup$
add a comment |
$begingroup$
It is easy to get
begin{equation}
10 = frac{x}{tan 23^{o}}-frac{x}{tan 32^{o}}
end{equation}
and then you can obtain that
begin{equation}
x = frac{10}{frac{1}{tan 23^{o}} - frac{1}{tan 32^{o}}}
end{equation}
answered Nov 17 '14 at 3:51
askuyueaskuyue
16510
$endgroup$
add a comment |
$begingroup$
It is easy to get
begin{equation}
10 = frac{x}{tan 23^{o}}-frac{x}{tan 32^{o}}
end{equation}
and then you can obtain that
begin{equation}
x = frac{10}{frac{1}{tan 23^{o}} - frac{1}{tan 32^{o}}}
end{equation}
answered Nov 17 '14 at 3:51
askuyueaskuyue
16510
$endgroup$
It is easy to get
begin{equation}
10 = frac{x}{tan 23^{o}}-frac{x}{tan 32^{o}}
end{equation}
and then you can obtain that
begin{equation}
x = frac{10}{frac{1}{tan 23^{o}} - frac{1}{tan 32^{o}}}
end{equation}
answered Nov 17 '14 at 3:51
askuyueaskuyue
16510
answered Nov 17 '14 at 3:51
askuyueaskuyue
16510
answered Nov 17 '14 at 3:51
askuyueaskuyue
16510
answered Nov 17 '14 at 3:51
askuyueaskuyue
16510
16510
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1023958%2fmath-question-angle-of-elevation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Let QR = y. Can you use the two right angled triangles? Use them to set up two equations and solve them for y and x.
$endgroup$
– Mick
Nov 16 '14 at 10:05
$begingroup$
I am using tan32 = x/y +> y = x/tan32 and for x: tan 23 = x/10+y
$endgroup$
– Asad Rafiq
Nov 16 '14 at 10:12