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39 votes
I need help on 8-9 plss :))

I need help on 8-9 plss :))-example-1
User Nitin Savant
by
2.8k points

1 Answer

18 votes
18 votes

Answer:

8. SU = 24

9. TU = 16√3

Explanation:

Recall: SOH CAH TOA

8. Reference angle (θ) = 30°

Opposite = 8√3

Adjacent = SU

Apply TOA,

Tan θ = Opp/Adj

Substitute

Tan 30° = 8√3/SU

Tan 30° × SU = 8√3

SU = 8√3/Tan 30°

SU = 8√3/(1/√3) (tan 30° = 1/√3)

SU = 8√3*√3/1

SU = 8*3

SU = 24

9. Reference angle (θ) = 30°

Opposite = 8√3

Hypotenuse = TU

Apply SOH,

Sin θ = Opp/Hyp

Substitute

Sin 30° = 8√3/TU

Sin 30° × TU = 8√3

TU = 8√3/sin 30°

TU = 8√3/(½) (sin 30° = ½)

TU = 8√3 × 2/1

TU = 16√3

User Himanshu Shekhar
by
3.2k points
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