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y=24.5^(∘)
y=7.2
Notice that the central angle ∠BCD has the same measure as its intercepted arc BD. Let's add that information to the diagram.
Using the Triangle Angle Sum Theorem, we can write an equation containing x. m∠x+m∠x+106^(∘)=180^(∘) Let's solve this equation by performing inverse operations.
Since the two sides that are cut by transversals are parallel, we know by the Consecutive Interior Angles Theorem that they are supplementary. With this information we can write two equations.
Let's solve the second equation as well.
Substitute values
LHS * sin 57^(∘)=RHS* sin 57^(∘)
a/c* b = a* b/c
Use a calculator
Round to 1 decimal place(s)
To find the length of y we must know the side's opposite angle. This will allow us to use the Law of Sines again. Since we know two angles in the triangle, we can use the Triangle Angle Sum Theorem to determine the remaining angle.
Add terms
LHS-130^(∘)=RHS-130^(∘)
Let's add this angle to the diagram.
Now we can use the Law of Sines again to solve for y.
Substitute values
LHS * sin 50^(∘)=RHS* sin 50^(∘)
a/c* b = a* b/c
Use a calculator
Round to 1 decimal place(s)
Using the Triangle Angle Sum Theorem we can write and solve an equation containing x.
Remove parentheses
Add and subtract terms
LHS+20^(∘)=RHS+20^(∘)
.LHS /20.=.RHS /20.