Question

In the adjacent figure ∆ MNP is a right angled triangle
and angle MNP = 90, angle MPN = and sin ө = 5/13
Find cos ө and tan ө​

Answer 1

cos ө = 12/13, tan ө = 5/12.

Step-by-step explanation:

by pbp/hhb formula sin ө = perpendicular/hypoteneous, similarly, cos ө = base /hypoteneous, tan ө = perpendicular/ base. pytagoras formula given us h^2= p^2+b^2, so 13^2=p^2+ 12^2, then p^2 =169-144, p=√144, p= 12.so,answers are cosө =12/13, tan ө = 5/12.

Answer 2

GIVEN:-

angle MNP=90°

sin theta =15/13

TO FIND : -

COS THETA =?

tan theta = ?

SOLUTION:-

AS WE KNOW THAT,

SIN THETA = opposite/HYPOTENUSE

=15/13

therefore, opposite side of theta = MN = 15

and hypotenuse = MP=13

IN triangle MNP, MNP= 90°

BY PYTHAGORAS THEOREM,

(MP)^2=(PN)^2+(MN)^2

(13)^2= (PN)^2+(15)^2

169=(PN)^2+225

(PN)^2=225 - 169

= 56

therefore, PN = 7.4

cos theta = adjacent/ HYPOTENUSE

= 7.4/13

TAN THETA = OPPOSITE / ADJACENT

= 15/7.4.