1993 Sgs Q8 (1 Viewer)

[adzy]

The lengths of the sides of a triangle from an arithmetic progression and the largest angle of the triangle exceeds the smallest by 90 degrees. Find the ration of the lengths of the sides.

I dont want to trig bash anymore. Can anybody find an elegant solution?

 

[damo676767]

X² + (A + B)² = (A + 2B – X)²
X² + A² + 2AB + B² = A² + B² + X² + 4AB - 2AX -4BX
2AB - 2AX –4BX = 0
X(2A + 4B) = 2AB
X = AB/(A+2B)


Sin2@ = (A+B)/(A + 2B – X)


Sin@/X = Sin(180 – 2@) / A
= 2sin(90 - @) cos (90 - @) / A
= 2sin@cos@ / A
= sin2@/A

sin@ = X(A+B)/A(A + 2B – X)
= AB(A+B)/(A+2B)A(A+2B – AB/(A+2B))


sin@/(A + B) = sin(@ + 90)/(A + 2B)
= (sin@cos90 + sin 90 cos@) / (A + 2B)
= cos@ / (A + 2B)

cos @ = (A² + (A+2B) ² – (A + B) ²)/ 2A(A+2B)


AB/(A+2B)A(A+2B – AB/(A+2B)) = (A² + (A+2B) ² – (A + B) ²)/ 2A(A+2B)(A+2B)


from that it can be worked out, but its too dam hard when you are typing

see the picture attached for the positioning of my letters.