Help solving these two equations...?

Question

solve :

area = (1/2)a(225 - a) sin 90 and

area= (1/2)a(225-a)sin 70

Please explain how to complete these.

btw these are two seperate equations for two different angles, 70 & 90, they are not simultaneous equations

Thank you

Answer 1

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Answer 2

i am guessing you are solving for a triangle where two sides add up to 225

for the first one .. . .

this is a right triangle as you have a right angle (90°)

then let A = area ... i think you are solving for a = side

A = (1/2) a (225 - a) sin90

2A/sin90 = a(225 - a) .. . .. . but sin90 = 1

2A = 225a - a^2

a^2 - 225a + 2A = 0

use quadratic formula

a = [225 ± √(225^2 - 8A)]/2

choose a so that it is between 0 & 225

for the next one, replace 90 by 70

then

2A/sin70 = 225a - a^2

a^2 - 225a + 2A/sin70 = 0

a = [225 ± √(225^2 - {8A/sin70})]/2

0 < a < 225