the change for ₹ 3 is ______ 50 paise coins
Answers
Answer:
Find the area of the right angled triangle with hypothenuse 5 cm and one of the acute angle is 48°30'.
\purple{\large{\underline{\underline{ \rm{Solution: }}}}}
Solution:
Here, first we need to find AB and then BC. Therefore, by finding both height and breadth, we will apply the formula of area of right triangle.
From the figure,
For finding AB i.e, height (perpendicular)
\sf{ \sin C = \dfrac{perpendicular}{hypotenuse}}sinC=
hypotenuse
perpendicular
\sf \sin C = \dfrac{AB}{AC}sinC=
AC
AB
\sf{ \sin48 \degree30' = \dfrac{AB}{5}}sin48°30
′
=
5
AB
N.B :-
sin 48° = 0.7431
But we have sin 48°30'
Let's convert 30' into degrees, we have:
30' = 0.5°
Now 48° + 0.5° = 48.5°
∴ sin 48.5° = 0.7490
Let's continue.......
\sf{0.7490 = \dfrac{AB}{5}}0.7490=
5
AB
\sf5 \times 0.7490 = AB5×0.7490=AB5×0.7490=AB5×0.7490=AB
★\sf{AB = 3.7450 \: cm}AB=3.7450cm
For finding BC i.e, base
\sf \cos C = \dfrac{base}{hypotenuse}cosC=
hypotenuse
base
\sf{ \cos C = \dfrac{BC}{AC} }cosC=
AC