$\mbox{a rocket is shot straight into the air at an initial velocity of 96 feet/second. }$ $\mbox{the height of the rocket after
1 answer:
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Given:
Consider the below figure attached with this question.
∠EFH = (5x + 1)°, ∠HFG = 62°, and ∠EFG = (18x + 11)°
To find:
The measure of ∠EFH.
Solution:
From the figure it is clear that ∠EFG is divide in two parts ∠EFH and ∠HFG. So,
Isolate variable terms.
Divide both sides by 13.
The value of x is 4.
Therefore, the measure of ∠EFH is 21°.
