Question

How do i solve for a right triangle?

Answer 1

The confusion might have arrised by the fact that the angles given here are not the standard angles that we are familier with. The ones that we are taught in school.

But, one shall not be hesitated by this fact from moving further on solving the problem :)

Applying the trigonometry,

`sin34^o = (\text{SU})/9`

Checking in on the calculator, we get `sin 34^0 = 0.55`. Which can be predicted by the fact that `sin 30^0 = 0.50`. So, `sin 34^0` has to be a little bit greater than `0.5`.

Substituting the values we get `\text{SU} = 5.0`

I leave the 2nd part up to you.

For angle `S`,

`cos S = (\text{SU})/\text{TS}`

`cos S = 5.0/9`

`cos S = 0.55`

checking in the calculator we get,

`S = 57^0`

Answer 2

In a right triangle you can apply trigonometric functions.

Explanation:

`sin(34°)= (SU)/9 rArr SU=9*sin(34°)~~5`
`cos(34°)=(TU)/9 rArr TU=9*cos(34°)~~7.5`
Define angle `TSU=alpha`
`sin(alpha)=(TU)/9 => alpha=arcsin((TU)/9)=56.4`

For non-right triangle see: law of sines and law of cosines.

Answer 3

From top to bottom, `5.03, 7.46, and 56^o`

Explanation:

let's look at what we know:

We know `/_T`, and we know side length TS, which we know is the hypotenuse as it is directly opposite the right angle and we can see that it is the longest side.

First, let's find the side opposite to the angle, side length US.

Because we know an angle, and the hypotenuse, and we are trying to find the opposite side, or the side opposite to the angle, we should use SINE.

`"opposite"=sin(x)xx" hypotenuse"`, where `x` is the angle. ('hypotevse' is supposed to be hypotenuse, sorry!)

So,

`sin(34)xx 9 ≈ 5.03`

So we know side length US, now let's find TU.

We can use Pythagoras' Theorem here, or you could use either TAN or COS if you wanted to. I will use Pythagoras' Theorem as it is the simplest method.

`a^2 + b^2 = c^2`
`a^2 + 5.034^2 = 9^2`
`a^2 + 25.33 = 81`
`a^2 = 55.67`
`a = 7.46`

So side length `"TU" ≈ 7.46`

To find the angle, we know that there is 180° in all triangles, so we simply subtract the two angles we have from 180°

`180° - 34° - 90°`

`/_S=56^o`

Answer 4

SU = 5m
TU = 7.46m
`m < S = 56^o`

Explanation:

1. SU; `sin(theta) = ("opp")/("hyp")`

`color(white)("ddddd")sin (34)= ("SU")/9`

`color(white)("dd")9xx sin( 34)= "SU"`

`color(white)("ddddddd.d")"SU"= 9xx0.5592`

`color(white)("ddddddd.d")"SU"= 5m`

`color(white)("d")`

2 `"TU; " cos(theta) =("adj")/("hyp")`

`color(white)("ddddd")cos (34)= ("TU")/9`

`color(white)("dddddddd")"TU"= 9 xx cos(34)`

`color(white)("dddddddd")"TU"= 9xx 0.8290`
`color(white)("dddddddd")"TU"= 7.46m`

`color(white)("d")`

3 `m < S =90^o -34^o =56^o`