Can you please tell how would would solve this quadratic question:-

The sides of a right-angled triangle are X, (x+2) and (2x-2). The hypotenuse is length (2x-2). Find the actual dimensions of the triangle.

and the Trigonometric Question:-
In triangle ABC, the angle A is 42 degrees, the side AB is 16cm and the side BC is 14cm. Find the two possible values of the side AC. How do you get two lengths?

:confused:

1. Use pythagorean theorem. The sum of the squares of each leg is the square of the hypotenuse.

(x)^2 + (x + 2)^2 = (2x - 2)^2

solve.

2. draw a triangle.

label each side and angle.

use the law of sines or law of cosines to find AC.

When I work multiply put and do everything I don't get the answer and for the next question how on earth get you get two possible answers?

What did you get for #1?

And did you try #2?

I understand the second one I figured it out its 2.88cm and 20.9cm, and for the first I still don't understand, can you please just show me how to to it?

What do you get when you multiply it out? I don't give out answers, I try to help you understand the material. Work it out step by step and show me what you get. If you have a specific question, ask away.

I get 9,10 and 13 which is wrong, honestly you've been of no help

To be honest, you've been pretty much just asking for the answer. Forum rules state that I am not to just give you the answer, but to help you work out the answer. If you had just simply worked it out, this would have been painless, but your lazy self simply didn't want to. Best of luck with other things in life.

I will show you how to do the first one. It is just a matter of solving the resulting quadratic. If I show you, then perhaps it'll help you on further problems? OK?

I assume you know the Pythagorean theorem. a^{2}+b^{2}=c^{2}

Your sides are a^{2}=x^{2}, \;\ b^{2}=(x+2)^{2}, \;\ c^{2}=(2x-2)^{2}

Now, sit it up:

x^{2}+(x+2)^{2}=(2x-2)^{2}

x^{2}+(x+2)^{2}-(2x-2)^{2}=0

Now, expand out using foil or whatever and you get:

-2x^{2}+12x=0

Can you solve this now? Use the quad formula or factoring.

Factor out a -2x and you should readily see the two solutions. Only one will be viable and that'll be your solution.

Thank you very much, I've realized the answers 6

and the Trigonometric Question:-
In triangle ABC, the angle A is 42 degrees, the side AB is 16cm and the side BC is 14cm. Find the two possible values of the side AC. How do you get two lengths?

:confused:

We can use the law of sines to find angle B and go from there.

\frac{14}{sin(42)}=\frac{16}{sin(C)}

sin(C)=\frac{16sin(42)}{14}

C=sin^{-1}(\frac{8sin(42)}{7})\approx{49.88}

Now, since we have C we can easily find angle B by just subtracting the two known angles from 180.

180-(42+49.88)=88.12

Now, use the law of cosines to find length AC. Can you do that OK?

(AC)=\sqrt{(BC)^{2}+(AB)^{2}-2(AB)(BC)cos(\text{angle B})}

Thank you so much you're a genius!

That would be "you're a genius". But thank you.:)

just out of curiosity, after looking at the above question can I just ask
for this question:
In triangle ABC, the angle A is 42 degrees, the side AB is 16cm and the side BC is 14cm. Find the two possible values of the side AC. How do you get two lengths?