G&T (geometry and trigonometry) grade/year 11 question...?

Question

I have a right-angled triangle. It is labelled A, B and C (for the different sides).
The trianglehas two angles labelled a and b as well as the 90degree angle.
a is 40degrees and is close to A. I was given values for the sides AB, AC and also BC for the previous question I completed. AB was 5, AC was 4 and BC was 3.

For the 1st part of this question (a) AB is 10 (double what is was in the 1st question, I don't know whether it holds any relevance.

The question:

Draw a separate new diagram of triangle ABC and label its dimensions for each of the following:

a) AB=10, AC=x, BC=y. Find x and y

b) AC=10, BC=z, AB=w. Find z and w

c) BC=10, AB=u, AC=v. Find u and v.

sin=O/H
cos=A/H
tan=O/A

O=Opposite
A=Adjacent
H=Hypotenuse.

Please help me! Please tell me how you did it. (write it down in the steps so I can follow so I can learn how to do it). Thankyou.

I should have mentioned that I have solutions to the 3 parts of the question but it doesn't make any sense to me. I'll tell you...

Solutions:
a)
x=10cos40degrees or 10sin50 degrees
=7.66 (rounded)

y=10sin40degrees or 10cos 50degrees

(for y of part a)
=6.43 (rounded)

question 1 part b)
b)
z=10tan40degrees
or 10/(divided by) tan 50 degrees
=8.39

w=10/cos40degrees or 10/sin50degrees
=13.05

c)
u=10/sin40degrees or 10/cos50degrees
=15.56

v=10/tan40degrees or 10tan50degrees
=11.92

Answer 1

Okay, so if I understand your question, you're working with different triangles that just have different size sides. But the angles are the same. As long as the sides grow proportionally, it doesn't matter how big the triangle gets, the angles are always going to be the same. I don't know if you learned this in Geo but they're called similar triangles.

Alright, now to start the problem, I would recommend finding the what each of the angles are. You know that two of the sides are 3 and 4, and the hypotenuse is 5. You already know that one of the angles is 90. So you just have so use those trig functions you have listed there to find the angle values. I'm just going to pick one... I'm going to just sin(theta) = opp/hypotenuse. One of the possibilities is.. sin(theta)= 3/5. When you put that into your calculator you have to type sin^(-1) (3/5). And that should give you an angle measure of 36.87 degrees. (make sure you're in degree mode) and you know that all the angles in a triangle have to add up to 180, so 180-(90+36.87) is 53.13. Which is the third angle.

Okay, now for part (a), you need to draw a triangle similar to the one in your previous problem, and label the angles you found, because remember the angles will be the same for every triangle. And then you can just trig again to find the other sides. So in your new triangle, the hypotenuse is 10, and you can find the other sides using either sin(theta)=opp/hyp or cos(theta)=adj/hyp so for one of the sides, you could write sin(53.13)=opp/10. And solve for that side. Which turns out to be 8. And then to solve for the third side, you could use cos(53.13)=adj/10. Which turns out to be 6. You just do the same thing for (b) and (c).

Answer 2

The first triangle they gave you was a 3, 4, 5 right triangle (also known as a Pythagorean Triple). Since you said AB was 5, that means C was the right triangle. I'm guessing that all the new triangles will be similar with C as their right triangle.

a) If AB = 10 here and AB was five in the last triangle then your ratio is 10/5. So to find your missing sides you would set up two proportions (one for each missing side).

For AC: 10/5 = x/4, notice the order I used was new triangle/old triangle for both ratios.

For BC: 10/5 = y/ 3.

b) Same idea different ratio. In this case, the new AC = 10 and the old AC was 4. So the ratio you would use here is 10/4.

for BC: 10/4 = z/3

You would just keep setting up ratios based on the sides measure they give you each time. You don't actually need any trig rations to solve these triangles.

Answer 3

just use proportions.
a)5/10=4/x, x=8 5/10=3/y, y=6
b)4/10=3/z, z=7.5 4/10=5/w, w=12.5
c)3/10=5/u, u=50/3 3/10=4/v, v=40/3
angles will always be the same.
sin cos and tan are not needed

Answer 4

particular... angles upload as much as a hundred and eighty ranges, and one perspective is ninety ranges; understanding another perspective provides the different perspective. something is what's called the sine rule: a/sin(A) = b/sin(B) = c/Sin(C) the place a is the dimensions of the area opposite perspective A b is the area length opposite perspective B c is the area length opposite perspective C (and, if C is the suited perspective, then sin(C) = a million....)