Simple Trig (1 Viewer)

nrlwinner

Member
Can someone help me because I just can't seem to get this.

A wooden stake, S, is 13m from a point, A, on a straight fence. SA makes an angle of 20degrees with the fence. If a goat is tethered to S by a 10m rope, where, on the fence, is the nearest point to A at which it can graze.

Dumbledore

Member
i can't really draw the diagram here and i'm not sure if i even got it right but

it looks like you have to use the cosine rule to find the distance from point A

u end up getting: 100 = x^2 + 169 - (26cos20)x but this quadratic is messy
solving this quadratic u get x ~= 3.26, u will get 2 solutions because there are two possible ways for the triangle to work, u take the smaller one since u want it closer to A

Alternatively

use the sine rule to find the angle between rope and fence,
u get @=26.4, 180-26.4(153.6)[since sine is + in first 2 quadrants and it can be either since they are both < 180]
next since it says u want the closest value to the fence, u use the optuse angle(153.6) and use the sum of triangle to figure out the angle between SA and the rope, u get 180-153.6-20 = 6.4 then use the sine rule once more and get x ~=3.26

i think ur supposed to use the 2nd method but both seem to get the same answer

nrlwinner

Member
i can't really draw the diagram here and i'm not sure if i even got it right but

it looks like you have to use the cosine rule to find the distance from point A

u end up getting: 100 = x^2 + 169 - (26cos20)x but this quadratic is messy
solving this quadratic u get x ~= 3.26, u will get 2 solutions because there are two possible ways for the triangle to work, u take the smaller one since u want it closer to A

Alternatively

use the sine rule to find the angle between rope and fence,
u get @=26.4, 180-26.4(153.6)[since sine is + in first 2 quadrants and it can be either since they are both < 180]
next since it says u want the closest value to the fence, u use the optuse angle(153.6) and use the sum of triangle to figure out the angle between SA and the rope, u get 180-153.6-20 = 6.4 then use the sine rule once more and get x ~=3.26

i think ur supposed to use the 2nd method but both seem to get the same answer
Yes you are right. Can someone put up a diagram because I can't get the what the actual question is saying.

jet

Banned

$\bg_white \text{The goat can graze at 2 points on the fence } G \text{ and } G'. \\ \text{Using the cosine rule } GS^2 = AS^2 + AG^2 - 2AS \cdot AG cos \, \angle SAG \\ 100 = 169 + x^2 - 26xcos20 \\ x^2 - \left (26cos \, 20^{\circ} \right )x + 69 = 0 \\ \text{The point } G \text{ will have the smaller value of } x \text{, which will also be the answer} \\ x = \frac{26cos \, 20^{\circ} - \sqrt{676cos^2 \, 20^{\circ} - 276}}{2} \\ = 13cos \, 20^{\circ} - \sqrt{169cos^2 \, 20^{\circ} - 69} \\ = 3.26 \text{ m}$

Galactic Drama

New Member
A bit late but where did you get this question from?

Drongoski

Well-Known Member
A bit late but where did you get this question from?
The poster most likely no longer active on Bored.

HazzRat

Active Member
A bit late but where did you get this question from?
Last post July 2009,..

tywebb

dangerman
A bit late but where did you get this question from?
It is from New Senior Mathematics Two Unit Course for Years 11 & 12 by Fitzpatrick. (1st ed, 1984) Ex 8d Q10 page 191

However it has been reused in the second and 3rd editions too.

In the more up-to-date 3rd edition it is Ex 2.8 Q10 on page 47 (slightly reworded, but essentially the same).

Galactic Drama

New Member
It is from New Senior Mathematics Two Unit Course for Years 11 & 12 by Fitzpatrick. (1st ed, 1984) Ex 8d Q10 page 191

However it has been reused in the second and 3rd editions too.

In the more up-to-date 3rd edition it is Ex 2.8 Q10 on page 47 (slightly reworded, but essentially the same).
Thankyou