Question

In triangle ABC ,AB= 8 cm,BC=5 cm and CA= 4 cm.find the greatest and smallest angle of triangle ABC.

Answer 1

Answer: greatest angle is C

   Smallest angle is B

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given that,

In ΔABC, AB=8cm, BC=5cm and CA=4cm

To find:

Greatest angle and Smallest angle in ΔABC.

So to find the angle's of the ΔABC, we have to use the SSS formula to find the angles.

∠$a = cos^{-}(\frac{b^{2} + c^{2} - a^{2} }{2bc})$

we have to make sure that when we put the values in "a", "b" and "c" we have to put the value opposite of the angle.

In this case,

Value of a = 5cm (BC)

Value of b = 4cm (CA) and

Value of c = =8cm (AB)

After substituting the values in the formula mentioned above,

∠$a = cos^{-}(\frac{b^{2} + c^{2} - a^{2} }{2bc})$

∠$a = cos^{-}( \frac{4^{2} + 8^{2} -5^{2} }{2*4*8} )$

∴ ∠a = 30°

Similarly after putting the values in the formula-

∠$b = cos^{-}(\frac{a^{2} + c^{2} - b^{2} }{2ac})$   and  ∠$c = cos^{-}(\frac{a^{2} + b^{2} - c^{2} }{2ab})$

we get

∠b = 24.14° ≅ 24° and  ∠c = 125°

So, according to the findings,

The greatest angle in ΔABC is ∠c and

The smallest angle in ΔABC is ∠b