Question

the change for ₹ 3 is ______ 50 paise coins​

Answer 1

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