9. El radio de una pelota de básquetbol es aproximadamente 3.6 veces mayor
1 answer:
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See the attached figure which represents the explanation of the problem.
Let <span>the height of the building = h
</span>
∵ ∠1 congruent to ∠2
∴ ∠1 = ∠2
∴ tan (∠1) = tan (∠2) ⇒⇒⇒ tan of the angle = (opposite/adjacent)
∴
∴ 12 h = 5 * 264 = 1,320
∴ h = 1,320/12 = 110 feet
∴ T<span>he height of the building = 110 feet
</span>
Let <span>the height of the building = h
</span>
∵ ∠1 congruent to ∠2
∴ ∠1 = ∠2
∴ tan (∠1) = tan (∠2) ⇒⇒⇒ tan of the angle = (opposite/adjacent)
∴
∴ 12 h = 5 * 264 = 1,320
∴ h = 1,320/12 = 110 feet
∴ T<span>he height of the building = 110 feet
</span>