Trig (1 Viewer)

Green Yoda · Jul 21, 2015

The answer is 970m and i got 46m lol

InteGrand · Jul 21, 2015

Rathin said:
View attachment 32228

The answer is 970m and i got 46m lol

$Draw a diagram as follows. Let $H$ be the point where the helicopter is hovering (500 m above the ground), and let $O$ be the point on the ground directly below $H$ (so $OH = 500$ (units of lengths everywhere in the diagram is metres)). On the ground, to the right (say) of $O$, draw a point $F$, the point where the fire starts at. Draw the point $G$ somewhere between $O$ and $F$, where $G$ is the new point the fire is at ($O$, $F$, and $G$ form a straight line as they are all points on the ground). Let $FG=x$. We want to find this length. Label $\angle HOF$ as a right angle. Draw a dotted line through $A$ parallel to the ground $OF$. Let $X$ be some point on this line on the same side of $HO$ as $F$. Then from the given data, $\angle XHF = 10^\circ$ and $\angle XHG = 15^\circ$.$

$Therefore, we can see from the diagram that $\angle FHG = 5^\circ$ (as $\angle FHG = \angle XHG - \angle XHF$ (adjacent angles)). Also, $\angle HFO = \angle XHF = 10^\circ$ (alternate angles, $HX \parallel OF$). So $\angle HGO = 15 ^\circ$ (external angle of $\triangle FHG$).$

$Now in right-angled triangle $FHO$, $\tan 10^\circ = \frac{500}{FO}\Rightarrow FO = \frac{500}{\tan 10^\circ}$. Similarly, from right-angled triangle $GHO$, we have $GO = \frac{500}{\tan 15 ^\circ}$ (as $\angle HGO = 15^\circ$).$

$\therefore x = FG = FO - GO$

$=\frac{500}{\tan 10^\circ}-\frac{500}{\tan 15^\circ}$

$$=969.6155.06....$ (by calculator)$

$$\approx 970$ (nearest integer).$

$Hence the answer is 970 metres, to the nearest metre.$

Green Yoda · Jul 21, 2015

InteGrand said:
$Draw a diagram as follows. Let $H$ be the point where the helicopter is hovering (500 m above the ground), and let $O$ be the point on the ground directly below $H$ (so $OH = 500$ (units of lengths everywhere in the diagram is metres)). On the ground, to the right (say) of $O$, draw a point $F$, the point where the fire starts at. Draw the point $G$ somewhere between $O$ and $F$, where $G$ is the new point the fire is at ($O$, $F$, and $G$ form a straight line as they are all points on the ground). Let $FG=x$. We want to find this length. Label $\angle HOF$ as a right angle. [B]Draw a dotted line through $A$ parallel to the ground $OF$. Let $X$ be some point on this line on the same side of $HO$ as $F$. Then from the given data, $\angle XHF = 10^\circ$ and $\angle XHG = 15^\circ$.$

$Therefore, we can see from the diagram that $\angle FHG = 5^\circ$ (as $\angle FHG = \angle XHG - \angle XHF$ (adjacent angles)). Also, $\angle HFO = \angle XHF = 10^\circ$ (alternate angles, $HX \parallel OF$). So $\angle HGO = 15 ^\circ$ (external angle of $\triangle FHG$).$
[/B]
$Now in right-angled triangle $FHO$, $\tan 10^\circ = \frac{500}{FO}\Rightarrow FO = \frac{500}{\tan 10^\circ}$. Similarly, from right-angled triangle $GHO$, we have $GO = \frac{500}{\tan 15 ^\circ}$ (as $\angle HGO = 15^\circ$).$

$\therefore x = FG = FO - GO$

$=\frac{500}{\tan 10^\circ}-\frac{500}{\tan 15^\circ}$

$$=969.6155.06....$ (by calculator)$

$$\approx 970$ (nearest integer).$

$Hence the answer is 970 metres, to the nearest metre.$

I didn't get it from this point.

Green Yoda · Jul 21, 2015

from draw a dotted line...

InteGrand · Jul 21, 2015

Rathin said:
I didn't get it from this point.

$Did you draw the diagram? The angle of depression is defined as $\angle XHF$ (when the fire is at $F$), and $\angle XHG$ (when the fire is at $G$).$

InteGrand · Jul 21, 2015

Rathin said:
from draw a dotted line...

Which part didn't you get?

InteGrand · Jul 21, 2015

Rathin said:
from draw a dotted line...

By dotted line, I meant like the dotted horizontal line in this picture: http://www.mathwords.com/a/a_assets/angle of depression.gif

Green Yoda · Jul 21, 2015

Yup i got it. Btw why is the angle of depression of the second sight of the fire angle XHG and not FHO. Thats the mistake i made.

InteGrand · Jul 21, 2015

Rathin said:
Yup i got it. Btw why is the angle of depression of the second sight of the fire angle XHG and not FHO. Thats the mistake i made.

$It cannot be $\angle FHO$ because of the definition angle of depression (the angle formed by the line joining the observer and the object with the horizontal). $\angle FHO$ does not describe this angle. When the fire is at $G$, point $F$ is irrelevant for the angle of depression now, since we are interested with an object that is at $G$, not $F$. So the angle of depression now will be $\angle XHG$, since this is the angle formed by the line segment joining the observer (at $H$) to the object (now at $G$, so this line segment is $HG$) and the horizontal ($XH$).$

Green Yoda · Jul 21, 2015

InteGrand said:
$It cannot be $\angle FHO$ because of the definition angle of depression (the angle formed by the line joining the observer and the object with the horizontal). $\angle FHO$ does not describe this angle. When the fire is at $G$, point $F$ is irrelevant for the angle of depression now, since we are interested with an object that is at $G$, not $F$. So the angle of depression now will be $\angle XHG$, since this is the angle formed by the line segment joining the observer (at $H$) to the object (now at $G$, so this line segment is $HG$) and the horizontal ($XH$).$

My teacher didnt explain properly for the angle of depression defination. What do you mean by the angle formed by the line joining the observer and the object with the horizontal?

InteGrand · Jul 21, 2015

Rathin said:
My teacher didnt explain properly for the angle of depression defination. What do you mean by the angle formed by the line joining the observer and the object with the horizontal?

I mean, consider the point where the observer is, and the object is. Connect these two points with a line. The angle this line makes with the horizontal is the angle of depression (this is for when the 'observer' is higher up than the 'object'). (I assume you know what I mean by angle formed by two lines.)

Green Yoda · Jul 21, 2015

I dont know if its my english ability but i just so fkin confused in these word problems

Flop21 · Jul 21, 2015

Rathin said:
I dont know if its my english ability but i just so fkin confused in these word problems

In questions like these I think it helps to immediately draw up a big and clear diagram after reading the question.

It can also help to draw another diagram of the 2 triangles within the 1st diagram separate from each other, with points still labeled correctly and any other info. Then you can easily see what trig you need to solve for x or whatever.

Trig (1 Viewer)

Green Yoda

Hi Φ

InteGrand

Well-Known Member

Green Yoda

Hi Φ

Green Yoda

Hi Φ

InteGrand

Well-Known Member

InteGrand

Well-Known Member

InteGrand

Well-Known Member

Green Yoda

Hi Φ

InteGrand

Well-Known Member

Green Yoda

Hi Φ

InteGrand

Well-Known Member

Green Yoda

Hi Φ

Flop21

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)