Question

I need Help with this question... the asnwer should not be a decimal... it's about special right triangles.

Answer 1

`\text{Answer: x = 13}\sqrt[]{6},\text{ y = }\frac{13\sqrt[]{6}}{2},\text{ z = 39}\sqrt[]{2}`

Below is the figure of the triangle

We have three unkown variables which are x, y, and z

Firstly, let us find the unknown side z

To find z, we will be applying SOH CAH TOA

The longest side is z

The opposite sides is 39

`\begin{gathered} \text{ applying SOH} \\ \text{ Sin }\theta\text{ = }\frac{opposite}{\text{hypotenus}} \\ \sin \text{ 45 = }(39)/(z) \\ \text{Introduce cross multiply} \\ z\text{ x sin 45 = 39} \\ \text{Divide both sides by sin 45} \\ \frac{z\cdot\text{ sin45}}{\sin\text{ 45}}\text{ = }\frac{39}{\sin \text{ 45}} \\ z\text{ = }\frac{39}{\sin \text{ 45}} \\ \text{ According to speciaal triangles; sin 45 = }\frac{\sqrt[]{2}}{2} \\ z\text{ = }\frac{39}{\frac{\sqrt[]{2}}{2}} \\ z\text{ = }\frac{39\text{ x 2}}{\sqrt[]{2}} \\ z\text{ = }\frac{78}{\sqrt[]{2}} \\ \text{Rationalize the expression} \\ z\text{ = }\frac{78\text{ x }\sqrt[\square]{2}}{\sqrt[]{2}\text{ x }\sqrt[]{2}} \\ z\text{ = }\frac{78\sqrt[]{2}}{2} \\ z\text{ = 39}\sqrt[]{2} \end{gathered}`

Find x

X can be find by applying the SOH CAH TOA

Let z = opposite

let x = adjacent

`\begin{gathered} \text{ Applying TOA} \\ \text{Tan}\theta\text{ = }\frac{opposite}{\text{adjacent}} \\ \text{opposite = z = 39}\sqrt[]{2} \\ x\text{ = adjacent} \\ \text{Tan 60 = }\frac{39\sqrt[]{2}}{x} \\ \text{Introduce cross multiply} \\ x\cdot\text{ tan 60 = 39}\sqrt[]{2} \\ \text{Divide both sides by tan 60} \\ \frac{x\cdot\text{ tan 60}}{\tan\text{ 60 }}\text{ = }\frac{39\sqrt[]{2}}{\tan\text{ 60}} \\ \text{According to special triangles}\colon\text{ Tan 60 }=\text{ }\sqrt[]{3} \\ x\text{ = }\frac{39\sqrt[]{2}}{\sqrt[]{3}} \\ \text{Rationalize the above surd} \\ x\text{ = }\frac{39\sqrt[]{2}\text{ x }\sqrt[]{3}}{\sqrt[]{3}\text{ x }\sqrt[]{3}} \\ x\text{ = }\frac{39\sqrt[]{6}}{3} \\ x\text{ = 13}\sqrt[]{6} \end{gathered}`

Find y

let y = adjacent

x = Hypotenus

Applying SOH, CAH TOA

`\begin{gathered} \text{ cos }\theta\text{ = }\frac{adjacent}{\text{Hypotenus}} \\ \text{Adjacent = y} \\ \text{Hypotenus = x = 13}\sqrt[]{6} \\ \text{Cos 60 = }\frac{y}{13\sqrt[]{6}} \\ \text{Cross multiply} \\ y\text{ = cos 6}0\text{ x 13}\sqrt[]{6} \\ \text{According to special angles : cos 60 = }(1)/(2) \\ y\text{= }(1)/(2)\text{ x 13}\sqrt[]{6} \\ y\text{ = }\frac{13\text{ x }\sqrt[]{6}}{2} \\ y\text{= }\frac{13\sqrt[]{6}}{2} \end{gathered}`