Step-by-step explanation:
✰ [tex] \underline{ \underline{ \large{ \tt{G\: I \: V \: E \: N}}}} : [/tex]
✰ [tex] \underline{ \underline{ \large{ \tt{T \: O \: \: F\: I \: N \: D}}}} : [/tex]
✰ [tex] \underline{ \underline{ \large{ \tt{S \: O
\: L \: U \: T\: I\: O \: N}}}} : [/tex]
In right angled triangle ABE :
Now , Using trigonometric ratio :
[tex] \large{ \tt{Sin \theta \: = \: \frac{Perpendicular}{Hypotenuse} }}[/tex]
Plug the known values :
⟶ [tex] \large{ \tt{ Sin(20) = \frac{h}{8} }}[/tex]
⟶ [tex] \large{ \tt{0.34 = \frac{h}{8}}} [/tex]
⟶ [tex] \large{ \tt{h = 0.34 \times 8}}[/tex]
⟶ [tex] \large{ \tt{h = 2.72}}[/tex]
Now , We have :
Finding the area of a parallelogram whose base and height are 15 and 2.72 respectively :
❀ [tex] \underline{ \underline{ \text{Area \: of \: parallelogram = \tt{b *\: h}}}}[/tex]
⟶ [tex] \large{ \tt{A = 15 *2.72}}[/tex]
⟶ [tex] \large{ \tt{A = 40.8 \: {units}^{2} }}[/tex]
Hence , Our final answer :
[tex] \large{ \boxed{ \sf{Height \: ( \: h \: ) = 2.72 \: units}}}[/tex]
[tex] \large{ \boxed{ \sf{Area \: of \: parallelogram \: ( \:A \: ) = \: 40.8 \: {units}^{2} }}}[/tex]
Hope I helped ! ♡
Have a wonderful day / night ! ツ
# [tex] \underbrace{ \tt{Carry \: On \: Learning}}[/tex] !! ✎
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