Question

In a
$\triangle \: ABC$
, is right angled at B ,if base line is AB =12 and BC = 5 then determine cos C .

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Answer 1

Answer:

given AB = 12 and BC = 5

then make a right angle triangle

so you know that cos = base / hypotenuse

perpendicular is given BC = 5

Base is given = AB = 12

applying phythagoras theorem

we get hypotenuse

H^2 = P^2 + B^2

H^2 = 144 + 25

H^2 = 169

H = √ 169

H = 13

you need cos c so measure from point c in this case base will be 5 cm and hypotenuse 13 cm

so cos c = base / hypotenuse

cos c = 5/ 13

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Answer 2

Answer:

$\huge\mathcal{\green{Hola!}}$

$\huge\mathfrak{\red{Answer}}$

$\huge\mathcal{\green{CosC= \frac{5}{13} }}$

Step-by-step explanation:

Given:-

• ∆ABC right angled at / B

• AB= 12

• BC = 5

To find:-

• Cos C

In rt / 'd ∆ ABC; using the Pythagoras theorem we have,

AC^2 = BC^2+ AB^2

$AC {}^{2} = {12}^{2} + {5}^{2}$

AC^2 = 144 + 25

AC^2= 169

$AC = \sqrt{169} = 13$

$\huge\mathcal{\green{Therefore,}}$

AC = 13 units

$\huge\mathcal{\green{Now,}}$

We know that;

Cos (theta) = Base/Hypotenuse

Base = BC = 5 units.

Perpendicular = AB = 12 units.

Hypotenuse = AC = 13 units.

$= > \cos(c) = \frac{5}{13}$

$\huge\mathcal{\green{All \ the \ very \ best!!}}$

$\huge\mathfrak{\red{@MissTranquil}}$

$\huge\fbox{\orange{be \ brainly}}$