# Solving for angles and sides

Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot

### Solving for angles and sides

11 and 13 I am having huge trouble with. 11 is: Joanne has to replace the two supporting guy wires for a hydro pole. She measures the distance between the base of the wires to be 10m, and their angle of inclination to be 50° and 35°. Determine the total length of the guy wire that needs to be replaced. Determine the height of the pole.

Question 13 is: From his nest, Mr. Robin flies east for 80m then south for 60m and then straight up into the sky for 30m. At this point how far is Mr. Robin from his nest? And at what angle of elevation viewed from their nest does Mrs. Robin see Mr. Robin

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### Re: Solving for angles and sides

For 12 a)

The key is in solving the scalene triangle first, you have 4 measurements for it if you look closely. You know that the shorter guy wire has an angle of 50degress on the acute side, so the obtuse side will have and angle of 130degrees. Then you have all the angles you need, you can figure out the third by taking 130 and 30 (160) from 180 (20degrees).Your 4 measurements for the scalene triangle will therefore be 130degrees, 30degrees, 20degrees and 10cm. Then use the law of sines, a/Sin(A) = b/Sin(B) = c/Sin(C), to figure out the lengths of the sides. In this instance your equation would be 10/Sin20 = b/Sin130 = c/Sin30:
Then it's just algebra figuring out each side.

For 12 b)

You're using the measurements you learned from 12 a) in the law of sines again, figure out the angles in the right angled triangle, 2 are already there, and then use the side you already figured out in 12 a) to solve the problem.

For 13 a)

This is just Pythagoras, a2 + b2 = c2. Although you're thinking in 3 dimensions, it's the same principle. East 80cm, South 60cm, that's a right angle triangle, solve the hypotenuse. Then for the height, your measurements are 60cm and 30cm, find the hypotenuse. Add the hypothesis together and you have your answer.

For 13 b)

This question requires the use of the law of cosines, c^2 = a^2 + b^2 − 2ab cos(C). I'd suggest looking tat one up as it's slightly more complicated!

Masheroni

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### Re: Solving for angles and sides

Ignore my solution for 13 a) !!!!!
You'd have to use the first hypotenuse to figure out the OTHER triangle! Sorry!

Masheroni

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### Re: Solving for angles and sides

Good afternoon!

11)
h = height of the pole
g1 = length of guy wire (50)
g2 = length of guy wire (35)
x = distance between pole and guy wire g1 (50)
x+10 = distance between pole and guy wire g2 (35)

So:
$$\tan 50^{\circ}=\dfrac{h}{x}\\\cot 50^{\circ}=\dfrac{x}{h}\\x=h\cot 50^{\circ}\\ \tan 35^{\circ}=\dfrac{h}{x+10}\\\cot 35^{\circ}=\dfrac{x+10}{h}\\x+10=h\cot 35^{\circ}\\ h\cot 35^{\circ}=x+10=h\cot 50^{\circ}\\ h\cot 35^{\circ}-h\cot 50^{\circ}=10\\ h=\dfrac{10}{\cot 35^{\circ}-\cot 50^{\circ}}\\ h\approx 17,0$$

Now, you have the length of the pole, the guy wire is easy
$$\sin 50^{\circ}=\dfrac{h}{g_1}\\g_1=\dfrac{17}{\sin 50^{\circ}}\approx 22,2$$
$$\sin 35^{\circ}=\dfrac{h}{g_2}\\g_2=\dfrac{17}{\sin 35^{\circ}}\approx 29,6$$
Total length:
$$22,2+29,6=51,8$$

Let's solve the others in another post

Baltuilhe

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### Re: Solving for angles and sides

Good afternoon (2)!

13)

Going to east, 80m and to south, 60m, using Pythagoras:
$$d^2=80^2+60^2=6\,400+3\,600\\d^2=10\,000\\d=100m$$

And now, up 30m.
$$d^2=100^2+30^2=10\,000+900=10\,900\\d=\sqrt{10\,900}\approx 104,40m$$

Now, the angle of elevation:
$$\tan\theta=\dfrac{30}{100}=0,3\\\theta=\arctan 0,3\\\theta\approx 16,70^{\circ}$$

I hope I have helped!

Baltuilhe

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Joined: Fri Dec 14, 2018 3:55 pm
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