A target has a bull’s-eye with a diameter of 2 inches. The outer ring is 1 inch wide. What is the area of the outer ring? A circ
2 answers:
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Answer:
The angle F = 47.25°
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that f =42inches and e = 39 inches
Given that ∠E = 43°
By using sine rule
⇒
Cross multiplication , we get
⇒ 39 sin F = 42 × sin 43°
⇒ sin F = 0.734459
<u><em>Step(ii):-</em></u>
⇒ sin⁻¹ (sin F)= sin ⁻¹ ( 0.734459)
⇒ F = 47.25
The angle F = 47.25°