Question

if triangle abc is right angled at c ab= 13 and bc = 5 then find the value of sin A and cos A ​

Answer 1

Answer:

5/13,12/13

Step-by-step explanation:

first if you will see the triangle you will observe perpendicular was missing use Pythagoras theorum and your answer will be 12,

then , you will see you have all the three hypotenuse , perpendicular and base as

H=13,

P=12

B=5

now sin A = P/H (formula of sin ) =5/15

cos A = B/H ( formula of cos) =12/13

Answer 2

Answer:

$\huge\star \underline{ \boxed{ \purple{Answer}}}\star$

$\sf{sin\:A\:=\:\frac{5}{13}}$

$\sf{cos\:A\:=\:\frac{12}{13}}$

Step-by-step explanation:

Solution :-

In △ABC

By pythagorus theorem,

$\sf{(AB)^2\:=\:(AC)^2\:+\:(BC)^2}$

putting values,

➝$\sf{(13)^2\:=\:(AC)^2\:+\:(5)^2}$

➝$\sf{(13)^2\:-\:(5)^2\:=\:(AC)^2}$

➝$\sf{169\:-\:25\:=\:(AC)^2}$

➝$\sf{144\:=\:(AC)^2}$

➝$\sf{\sqrt{144}\:=\:AC}$

➝$\sf{12\:=\:AC}$

So the ratios of ∠A are,

Base = AC = 12

Perpendicular = BC = 5

Hypotenuse = AB = 13

∴ sin A = $\sf{\frac{Perpendicular}{Hypotenuse}}$

➝sin A = $\sf{\frac{5}{13}}$

& cos A = $\sf{\frac{Base}{Hypotenuse}}$

➝cos A = $\sf{\frac{12}{13}}$

Hope it helps :)