3 years ago

If Sinx= 3/4, find x (3sf)

1 answer:

3 0

I don't know what (3sf) means

If sine(x) = 3/4
Then you want to know the angle whose sine = 3/4.
This is called the arc sine.
arc sine (3/4) = 48.59 degrees

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12. In the given figure, RS is parallel to PQ, If RS = 3 cm, PQ = 6 cm and ar(∆TRS) = 15cm³, then ar (∆TPQ) = ? (a) 70 cm² (b) 5

Gnesinka [82]

$\large\underline{\sf{Solution-}}$

Given that,

In triangle TPQ, 

RS || PQ,

RS = 3 cm,

PQ = 6 cm,

ar(∆ TRS) = 15 sq. cm

As it is given that, RS || PQ

So, it means

⇛∠TRS = ∠TPQ [ Corresponding angles ]

⇛ ∠TSR = ∠TPQ [ Corresponding angles ]

$\rm\implies \: \triangle TPQ \: \sim \: \triangle TRS \: \: \: \: \: \: \{AA \}$

Now, We know 

Area Ratio Theorem,

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{ar( \triangle \: TRS)} = \dfrac{ {PQ}^{2} }{ {RS}^{2} }$

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = \dfrac{ {6}^{2} }{ {3}^{2} }$

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = \dfrac{36 }{9}$

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = 4$

$\rm\implies \:ar( \triangle \: TPQ) = 60 \: {cm}^{2}$

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