Question

plzzz help this question please​

Answer 1

Step-by-step explanation:

Using \: Pythagoras \: theorem \: , we \: have \: BC2=AB2+CA2⇒BC=62+62−−−−−−√ ⇒BC=62–√ cm \: BC2=AB2+CA2⇒BC=62+62 ⇒BC=62 cm Thus \: , d \: (B, C) = 62–√62 cm (4) In \: right \: ∆ABC \: , AB = \: CA \: = 6 cm ∴ ∠ACB \: = \: ∠ABC \: (Equal \: sides \: have \: equal \: angles \: opposite \: to \: them) \: Also, \: ∠ACB \: + ∠ABC \: = 90º (Using \: angle \: sum \: property \: of \: triangle) ∴ 2∠ABC \: = 90º \: ⇒ ∠ABC \: = 90 \: °290 \: °2 \: = 45º \: Thus, \: the \: measure \: of \: ∠ABC \: is \: 45º.UsingPythagorastheorem,we have BC2=AB2+CA2⇒BC=62+62−−−−−−√ ⇒BC=62–√ cmBC2=AB2+CA2⇒BC=62+62 ⇒BC=62 cm. Thus, d(B,C)= 62–√62 cm(4)In Right ∆ABC,AB=CA=6cm∴ ∠ACB=∠ABC (Equal sides have equal angles opposite to them)Also, ∠ACB+ ∠ABC=90º (Using angle sum property of triangle)∴2∠ABC =90º⇒ ∠ABC= 90°290°2 =45ºThus,the measured ∠ABCis 45º.

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Answer 2

Step-by-step explanation:

Question : Prove that√5 is irrational.

Answer :

Let us assume that √5 is a rational number.

Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒√5=p/q

On squaring both the sides we get,

⇒5=p²/q²

⇒5q²=p² —————–(i)

p²/5= q²

So 5 divides p

p is a multiple of 5

⇒p=5m

⇒p²=25m² ————-(ii)

From equations (i) and (ii), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5

⇒q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√5 is an irrational number

Hence proved