Question

If angle A of a triangle ABC is doubled and lengths of sides AB and AC are kept the same, the area of triangle remains the same. Then angle A =
a) 30° b) 45°
c) 60° d) 75°​

Answer 1

for the given situation, both the values of angle should sum up to 90⁰ with a difference of 30⁰

so answer is 30⁰

pls mark as brainliest

Answer 2

solution-

Then area of triangle =s(s−a)(s−b)(s−c)where s=21(a+b+c)

If sides are doubled the new sides 2a,2b,2c

Then S=21(2a+2b+2c)⇒2×21(a+b+c)=2s

Then area of new triangle =$$\sqrt{S(S-2a)(S-2b)(S-2c)}= \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{16}s(s-a)(s-b)(s-c)= 4\sqrt{s(s-a)(s-b)(s-c)}

Then area of new triangle =4 times of old triangle