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What is the measure of ∠XBC?

1. m∠XBC = m∠BAC + m∠BCA

2. 3p – 6 = p + 4 + 84

3. 3p – 6 = p + 88

4. 2p – 6 = 88

5. 2p = 94

m∠XBC =

What is the measure of XBC 1 mXBC mBAC mBCA 2 3p 6 p 4 84 3 3p 6 p 88 4 2p 6 88 5 2p 94 mXBC class=

Respuesta :

calculista
we Know that
m∠ABC=180-[(p+4)+84]--------> 180-[p+88]=92-p
m∠ABC=(92-p)°    equation 1
and
m∠ABC+(3p-6)=180----------m∠ABC=180+6-3p
 m∠ABC=186-3p   equation 2
(1)=(2)
(92-p)=186-3p-----------> 186-92=-p+3p--------------> 2p=94

p=47°
therefore
m∠XBC=(3p-6)°---------->(3*47-6)=135°

the answer is 
m∠XBC=135°
frika

The measure of an exterior angle of triangle is equal to the sum of measures of two interior angles of triangle that are not supplementary with this exterior angles.

In the case of this question this fact sounds as

m∠XBC = m∠BAC + m∠BCA (option 1).

Now if

  • m∠XBC=(3p-6)°
  • m∠BAC=(p+4)°
  • m∠BCA=84°,

then

3p-6=p+4+84 (option 2),

3p-6=p+88 (option 3),

3p-p-6=p-p+88,

2p-6=88 (option 4),

2p-6+6=88+6,

2p=94 (option 5),

p=47.

Then m∠XBC=(3p-6)°=(3·47-6)°=135°.

Answer: m∠XBC=135°.

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