¿Sera cierto que existe una fraccion con numerador 20 equivalente a la fraccion 100/55?Justifica tu respuesta
1 answer:
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The picture in the attached figure
step 1
we know that
It is given that AD and BD are bisectors of ∠CAB and ∠CBA respectively.
Therefore,
x = ∠CAB/2 -----> equation 1
y = ∠CBA/2 -----> equation 2
step 2
In triangle ABC,
∠CAB + ∠CBA + ∠ACB = 180° ----> [The sum of all three angles of a triangle is 180°]
∠CAB + ∠CBA + 110° = 180°
∠CAB + ∠CBA = 180° - 110°
∠CAB + ∠CBA = 70° ------> divide by 2 both sides
∠CAB/2 + ∠CBA/2 = 70/2 -------> equation 3
substitute equation 1 and equation 2 in equation 3
x+y=35
hence
the answer is
x+y =35°
⇒ x + y = 35° ...[From equation (1) and (2)]
step 1
we know that
It is given that AD and BD are bisectors of ∠CAB and ∠CBA respectively.
Therefore,
x = ∠CAB/2 -----> equation 1
y = ∠CBA/2 -----> equation 2
step 2
In triangle ABC,
∠CAB + ∠CBA + ∠ACB = 180° ----> [The sum of all three angles of a triangle is 180°]
∠CAB + ∠CBA + 110° = 180°
∠CAB + ∠CBA = 180° - 110°
∠CAB + ∠CBA = 70° ------> divide by 2 both sides
∠CAB/2 + ∠CBA/2 = 70/2 -------> equation 3
substitute equation 1 and equation 2 in equation 3
x+y=35
hence
the answer is
x+y =35°
⇒ x + y = 35° ...[From equation (1) and (2)]