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Lesechka [4]
2 years ago
7

Hi i know im desperate but pls ill give you BIG BIG points…

Mathematics
2 answers:
lorasvet [3.4K]2 years ago
7 0

m∠ABC

∠A, first off we can use vertical angles to figure out that angle A is 90-31 which is 59° and since we know C is 40° we can make an equation

59+40+x=180 which we get that angle B equals 81° which makes m∠ABC Acute

m∠CAB

this is an acute triangle because it is the exact same as m∠ABC, but just expressed differently.

Zina [86]2 years ago
4 0
Answer:
mABC = 81°
mCAB = 59°

Let’s start with finding mCAB.
Segment CD is a straight line that runs through A, so we know it is a total of 180°.
The problem tells us that there is a right angle (90°) and gives us that 31° in between. Since they all have to add up to 180, we can do
180 - 90 - 31 = ...59°!

Now that we have mCAB, we can find mABC.
All the angles in any triangle must add up to 180. Since we now know mCAB (59°) and mCBA (40°), we can subtract to find the missing angle.
180 - 59 - 40 = ...81°!

Hope this helps! Brainliest always appreciated
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Step-by-step explanation:

Let the 1st part of your answer be x , so the 2nd part will be 40-x . From the given information, we can write the equation: (1/4)x = (3/8) × (40-x) . We can simplify this into (1/4)x = (120-3x)/8 ; 8x = 480-12x ; 8x+12x = 480 ; 20x = 480 ; x = 480/20; x = 24

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3 years ago
Consider the following sets:
bagirrra123 [75]
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2. In the circle shown, secants PE and PF have been drawn such that mCE  82 , mDF  98 and the ratio 3. In circle O, tangent
Andrew [12]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The arc PR is  \theta _{PR}  = 84

Step-by-step explanation:

A descriptive diagram of the diagram given in the question is shown on the first uploaded image

  In the circle tangent  \=  {M \= Q} , secant PE and  chord PF are drawn

given that  \theta _ {MPQ} = 102^o , \theta _{M} =  60^o

   The objective is to determine PR

From the diagram we see that MQ is a straight which implies that the the angle

      \theta_{NPM}  =  180 -102

      \theta_{NPM}  =  78^o

Now looking at triangle NMP we see that

          \theta _{MNP} =  180  -[ \theta_{NMP} + \theta _{NPM}]

         \theta _{MNP} =  180  - [60 + 78 ]

         \theta _{MNP} =  42 ^o

Now the measure of an inscribed angle is half the measure of its  arc intercepted, this statement is the inscribed angle theorem

 So with the knowledge

   Then

           \theta_{RNP} =  \frac{1}{2}  \theta_{PR}

  Looking at the diagram we see that

           \theta _{MNP} = \theta_{RNP} = 42 ^o

             42 = \frac{1}{2}  \theta _{PR}

              \theta _{PR}  = 2  *  42

               \theta _{PR}  = 84

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3 years ago
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