Question

If in a ∆ABC,r3=r1+r2+r, then angle A +angle B =
a) 120°
b) 100°
c) 90°
d) 80°​

Answer 1

∠A+∠B = 90°

Step-by-step explanation:

In the given triangle ABC,

$r_3=r_1+r_2+r$....(1)

To find, value of ∠A+∠B

From equation (1),

$r_3-r=r_1+r_2$

We know that,$length=\frac{arc}{radius}$

So,$\frac{\Delta}{s-c}-\frac{\Delta}{s}=\frac{\Delta}{s-a}+\frac{\Delta}{s-b}$

⇒$\Delta(\frac{1}{s-c}-\frac{1}{s})=\Delta (\frac{1}{s-a}+\frac{1}{s-b})$

⇒$\frac{s-s+c}{(s-c)s}=\frac{s-b+s-a}{(s-a)(s-b)}$

⇒$\frac{c}{(s-c)s}=\frac{2s-(b+a)}{(s-a)(s-b)}$

here $s = semi\ perimeter = \frac{a+b+c}{2}$

⇒2s=a+b+c

⇒2s-(a+b)=c

Now,$\frac{c}{(s-c)s}=\frac{c}{(s-a)(s-b)}$

⇒(s-c)s=(s-a)(s-b)

⇒$s^2-cs=s^2-bs-as+ab$

⇒s(a+b-c)=ab

⇒(a+b+c)(a+b-c)=2ab

⇒$(a+b)^2-c^2=2ab$

⇒$a^2+b^2+2ab-c^2=2ab$

⇒$a^2+b^2=c^2$

This equation comes from Pythagoras Theorem

Hence ΔABC is right angle triangle,where ∠C=90°

Also,∠A+∠B+∠C=180°  [angle sum property of triangle]

       ∠A+∠B+90°=180°

        ∠A+∠B=180°-90°

        ∠A+∠B = 90°

Answer 2

Answer:90⁰

Step-by-step explanation: