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Notice that we can use one of the trigonometric ratios to evaluate the value of x. Let's recall that the tangent of ∠A is the ratio of the leg opposite ∠A to the leg adjacent ∠A. Using this definition, we can create an equation for tan 50^(∘).
LHS * 14=RHS* 14
Rearrange equation
Use a calculator
Round to 1 decimal place(s)
The value of x is approximately 16.7 feet. To find the distance between the ends of the class we should multiply x by two. 16.7* 2=33.4 The distance between the ends of the class is approximately 33.4 feet. Notice that this value is only an approximation, as we used approximate values to evaluate it.
To find the value of y we can use one of the trigonometric ratios. Again recall that the tangent of ∠A is the ratio of the leg opposite ∠A to the leg adjacent ∠A. Using this definition, we can create an equation for tan (50^(∘)+10^(∘)).
tan (50^(∘)+10^(∘))=y/14
Add terms
LHS * 14=RHS* 14
Rearrange equation
Use a calculator
Round to 1 decimal place(s)
The value of y is approximately 24.2 feet. Let's add this information to our diagram.
Now we will evaluate the difference of the distances between the center and the end of the class. To do this we will subtract 16.7 from 24.2. 24.2- 16.7=7.5 Since we are given that each student needs 2 feet of space we will divide 7.5 by 2 and round the result to the nearest integer that is not greater than this number. 7.5/2=3.75≈ 3 After increasing the angle, approximately 3 more students can fit at the end of each row.