Question

nly)
d) solve for x, y, z
triangle with angles 30°, 60°, sides 8, z, y, x

Explanation:

Step1: Find x using cosine

In the right triangle with hypotenuse 8 and angle 60°, $\cos60^\circ=\frac{x}{8}$. Since $\cos60^\circ=\frac{1}{2}$, $x=8\times\frac{1}{2}=4$.

Step2: Find height (let it be h) using sine

$\sin60^\circ=\frac{h}{8}$. $\sin60^\circ=\frac{\sqrt{3}}{2}$, so $h=8\times\frac{\sqrt{3}}{2}=4\sqrt{3}$.

Step3: Find y using tangent

In the left right triangle with angle 30° and height $4\sqrt{3}$, $\tan30^\circ=\frac{h}{y}$. $\tan30^\circ=\frac{1}{\sqrt{3}}$, so $y=h\sqrt{3}=4\sqrt{3}\times\sqrt{3}=12$.

Step4: Find z using cosine

$\cos30^\circ=\frac{h}{z}$. $\cos30^\circ=\frac{\sqrt{3}}{2}$, so $z=\frac{h}{\cos30^\circ}=\frac{4\sqrt{3}}{\frac{\sqrt{3}}{2}}=8$. (Note: Correction—wait, no: left triangle hypotenuse z: $\sin30^\circ=\frac{h}{z}$? No, angle at left vertex is 30°, so opposite side is h. So $\sin30^\circ=\frac{h}{z}$ → $z=\frac{h}{\sin30^\circ}=\frac{4\sqrt{3}}{0.5}=8\sqrt{3}$. Yes, earlier mistake fixed: angle 30° opposite side is height h, so $\sin30°=h/z$ → z=h/sin30°=8√3. And y: adjacent side to 30° is y, so $\tan30°=h/y$ → y=h/tan30°=4√3/(1/√3)=12. Correct.)

Answer:

$x = 4$, $y = 12$, $z = 8\sqrt{3}$