Question

in triangle ABC, Â = 14.3° , BC = 14 m. find AC.
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Answer 1

Step-by-step explanation:

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In ∆ABC, AB = 15 cm , BC = 13 cm and AC = 14 cm.

↪ By Heron's formula ,

S = ( A + B + C ) / 2S=(A+B+C)/2

S = ( 15 + 13 + 14 ) / 2S=(15+13+14)/2

S = 42 / 2S=42/2

S = 21S=21

↪ Now, area\:of\:triangle:-Now,areaoftriangle:−

= √S (S–A) (S–B) (S–C)

= √21 (21–15) (21–13) (21–14)

= √21 (6) (8) (7)

= √7056

= 84 cm^2=84cm

2

↪ Again,area\:of\:triangle,Again,areaoftriangle,

1/2×base×altitude=84 cm^21/2×base×altitude=84cm

2

1/2×14×altitude=841/2×14×altitude=84

7×altitude=847×altitude=84

altitude=84/7altitude=84/7

Altitude=12cmAltitude=12cm

Therefore, the area of ∆ABC is 84cm² and its altitude on AC is 12 cm.

Answer 2

Step-by-step explanation:

14.3+14+ACequal to 180

151.7ans

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