Question Number 149481 by mathdanisur last updated on 05/Aug/21

Answered by mr W last updated on 05/Aug/21

say AB=CK=x, BM=BK=1 AC=x−1+x=2x−1 (2x−1)^2 =x^2 +(x+1)^2 x^2 =3x ⇒x=3 ⇒AB=3,BC=4,CA=5 AK=(√(3^2 +1^2 ))=(√(10)) AK×AH=AM^2 (√(10))×AH=2^2 AH=((2(√(10)))/5) ⇒HK=(√(10))−((2(√(10)))/5)=((3(√(10)))/5) HC^2 =3^2 +(((3(√(10)))/5))^2 +2×3×((3(√(10)))/5)×(1/( (√(10)))) HC^2 =((81)/5) HC=(9/( (√5))) ((sin θ)/3)=(3/( (√(10))))×((√5)/9) sin θ=(1/( (√2))) ⇒θ=45°

Commented bymathdanisur last updated on 05/Aug/21

Thankyou Ser, cool

Commented byTawa11 last updated on 05/Aug/21

great sir

Commented byTawa11 last updated on 06/Aug/21

Sir mrW please check Q149510. Geometry. Your approach too is needed. Thanks sir for your time.

Commented bymr W last updated on 06/Aug/21

the question is answered. do you mean the answer is not correct?

Commented byTawa11 last updated on 06/Aug/21

No sir. Just to see another approach.

Commented byBrahan last updated on 06/Aug/21

why BK=BM=1

Commented bymr W last updated on 07/Aug/21

since we only need to get the angles, we don′t need to know the absolute lengthes of the sides, we only need to know the ratios between them. we can also take BK=BM=2 or any other values, or just take BK=BM=a.