Exercise 54 Page 594

From the figure, we see that ∠ KHL ≅ ∠ LHG. This means we can equate the expressions of these angles. ∠ KHL ≅ ∠ LHG: 5x^(∘)-27^(∘)=3x^(∘)+1^(∘)Since the equation only contains x as a variable, it can be solved.

5x^(∘)-27^(∘)=3x^(∘)+1^(∘)

SubEqn

LHS-3x^(∘)=RHS-3x^(∘)

2x^(∘)-27^(∘)=1^(∘)

AddEqn

LHS+27^(∘)=RHS+27^(∘)

2x^(∘)=28^(∘)

DivEqn

.LHS /2.=.RHS /2.

x^(∘)=14^(∘)

When we know that x^(∘)= 14^(∘), we can substitute this into the expression for ∠ KHL. ∠ KHL: 3( 14^(∘))+1^(∘)=43^(∘)