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Need some help with these 2 math questions, They are related so best to finish one then work on the bisect. Thank you.

Need some help with these 2 math questions, They are related so best to finish one-example-1
Need some help with these 2 math questions, They are related so best to finish one-example-1
Need some help with these 2 math questions, They are related so best to finish one-example-2
User Titiyoyo
by
8.2k points

1 Answer

4 votes

Answer:

∠RQT = 23.5

∠PQR = 133°

Explanation:

Question 1

Step 1 - Add angles together:

As all angles on a straight line must add up to 180°:

3x - 5 + x + 1 = 180

Add like terms together

3x + x - 5 + 1

4x - 4 = 180

Step 2 - Add 4 to both sides of the equation:

4x - 4 + 4 = 180 + 4

4x = 184

Step 3 - Divide both sides by 4:


(4x)/(4) = (184)/(4)

x = 46

∠RQS = x + 1

∠RQS = 46 + 1

∠RQS = 47

Given the ray QT bisects ∠RQS, the angles must be the same on either side so:

47 ÷ 2 = ∠RQT

∠RQT = 23.5

Question 2

As we know the value of x:

∠PQR = 3x - 5

Plug known values in:

3x - 5

3(46) - 5

138 - 5

∠PQR = 133°

This can also be calculated by subtracting known angle ∠RQS from 180:

180 - 47 = 133.

Hope this helps!

User Terry Ryan
by
8.3k points

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