Question

I need help with geometry. Were learning similarity and i have a test soon but im really confused. I've been trying to figure it out for the past 2 hours but I really have no idea. I attached a photo of my assignment from today if anyone could help me with that.

Answer 1

Consider the upper triangle

if we want to find b side, consider the following trigonometric identity:

`\cos \text{ (}\theta\text{) = }\frac{adjacent\text{ side}}{hypotenuse}`

in this case, we have:

`\cos \text{ (45) = }(b)/(20)`

solve for b:

`b\text{ = cos(45) x 20 = }\frac{\sqrt[]{2}}{2}\text{ x 20 = 10}\sqrt[]{2}`

then, we can conclude that

`b\text{ = 10}\sqrt[]{2}`

Now, for a-side, consider the following trigonometric identity:

`\sin \text{ (}\theta\text{) = }\frac{opposite\text{ side}}{hypotenuse}`

in this case, we have:

`\sin \text{ (}\theta\text{) = }(a)/(20)`

solve for a:

`a\text{ = }\sin \text{(45) x 20 = }\frac{\sqrt[]{2}}{2}\text{ x 20 = 10}\sqrt[]{2}`

then, we can conclude that

`a\text{ = 10}\sqrt[]{2}`

Now, for the c-side, consider the greater triangle :

if we denote the hypotenuse by h, then by Pythagorean theorem we have:

`h^2=20^2+15^2`

but h = a + c, then, replacing this in the previous equation we have

`h^2=(a+c)^2=20^2+15^2`

but, we know that a is

`a\text{ = 10}\sqrt[]{2}`

then we have:

`(10\sqrt[]{2}+c)^2=20^2+15^2\text{ = 625}`

now, taking the square root of both sides of the equation we obtain:

`10\sqrt[]{2}+c^{}=\text{ 25}`

solve for c:

`c^{}=\text{ 25}-\text{ 10}\sqrt[]{2\text{ }}\text{ }\approx10.85`

then we can conclude that :

`a\text{ = 10}\sqrt[]{2}`
`b\text{ = 10}\sqrt[]{2}`

and

`c^{}=\text{ 25}-\text{ 10}\sqrt[]{2\text{ }}\text{ }\approx10.85`