Question

$\huge\underline\mathfrak\blue{QUESTION}$
____________________________________
In △ABC, AB = 9cm, BC = 11cm, AC = 15cm. Find the smallest angle of △ABC.
____________________________________
$\large \mathbb \green{SOLVE \: IT \: DON'T \: SPAM!! \: }$
IF U KNOW THE RIGHT ANSWER THEN ONLY GIVE THE ANSWER!!
WRONG ANSWERS AND SPAMMERS WILL BE REPORTED!!
URGENT!!
$\large \underline \mathbb \gray{itzURGENT!!}$
​

Answer 1

Step-by-step explanation:

Here ABC is a triangle in which

AB = 15 cm, AC = 15 cm and BC = 18 cm

Draw AD perpendicular to BC, D is mid-point of BC.

Then, BD = DC = 9 cm

in right angled triangle ABD,

By Pythagoras theorem, we get

AB

2

=AD

2

+BD

2

AD

2

=AB

2

−BD

2

AD

2

=(15)

2

−(9)

2

AD

2

=225−81

AD

2

=144

AD = 12 cm

(i) cos ∠ ABC = Base/ Hypotenuse

(In right angled ΔABD,∠ABC=∠ABD)

= BD / AB

= 9/15

= 3/5

(ii) sin ∠ACB=sin∠ACD

= perpendicular/ Hypotenuse

= AD/AC

= 12/15

= 4/5