Answer:
a = 7; b = ⁷/2
Step-by-step explanation:
Recall the acronym, SOHCAHTOA
We would apply trigonometric ratio formula in finding a and b.
Reference angle [tex] (\theta) [/tex] = 60°
Opposite side to the reference angle = [tex] \frac{7\sqrt{3}}{2} [/tex]
Adjacent side = b
Hypotenuse = a
✔️To find a, we would apply SOH, which is:
[tex] sin (\theta) = \frac{opp}{hyp} [/tex]
Plug in the values
[tex] sin (60) = \frac{\frac{7\sqrt{3}{2}}{a} [/tex]
[tex] \frac{\sqrt{3}}{2} = \frac{\frac{7\sqrt{3}{2}}{a} [/tex] (sin 60 = √3/2)
[tex] \frac{\sqrt{3}}{2} = \frac{7\sqrt{3}{2}*\frac{1}{a} [/tex]
[tex] \frac{\sqrt{3}}{2} = \frac{7\sqrt{3}{2a} [/tex]
Cross multiply
[tex] \sqrt{3}*2a = 7\sqrt{3}*2 [/tex]
[tex] 2a\sqrt{3} = 14\sqrt{3} [/tex]
Side both sides by √3
[tex] 2a = 14 [/tex]
Divide both sides by 2
a = 14/2
a = 7
✔️To find b, apply TOA, which is:
[tex] tan (\theta) = \frac{opp}{adj} [/tex]
Plug in the values
[tex] tan (60) = \frac{\frac{7\sqrt{3}{2}}{b} [/tex]
[tex] \sqrt{3} = \frac{\frac{7\sqrt{3}{2}}{b} [/tex] (tan 60 = √3)
[tex] \sqrt{3} = \frac{7\sqrt{3}{2}*\frac{1}{a} [/tex]
[tex] \sqrt{3} = \frac{7\sqrt{3}{2b} [/tex]
Cross multiply
[tex] \sqrt{3}*2b = 7\sqrt{3} [/tex]
[tex] 2b\sqrt{3} = 7\sqrt{3} [/tex]
Divide both sides by √3
[tex] 2b = 7 [/tex]
Divide both sides by 2
b = 7/2
