Question

Please Solve this ! !!! I am giving many points for this question , Please Please solve this​

Answer 1

$\bf\large\red{\underline{given : }}$

• Angle PRQ = 60°
• PS = 20m
• △PQS is right angled at Q.
• PR = PS.

$\bf\large\red{\underline{ to \: find : }}$

• PQ = ?

$\bf\large\red{\underline{properties \: used : }}$

• Angle sum property :- sum of all angles in a triangle is 180°.
• Tregonometric theorems.
• Linear pair.
• Angle opposite to equal side is equal.

$\bf\large\red{\underline{ solution : }}$

☞ Angle PRQ and Angle PRS are supplementary.

∴ Angle PRQ + Angle PRS = 180° (linear pairs)

➜ 60 + Angle PRS = 180°

➜ Angle PRS = 120° ----------(1)

☞ In△PRS,

∵ PR = RS .

So, Angle RPS = Angle RSP. --------(2)

Angle RPS + Angle PRS + AngleRSP = 180°

(by angle sum property)

➜ 2 Angle PSR+ 120° = 180°

➜ 2 Angle PSR = 60°

➜ Angle PSR = 30° -------(3)

Angle PSR = Angle PSQ = 30°

(same angle of different triangle)

☞ In△PQS ,

sin PSQ = opposite/hypotenuse

➜sin PSQ = PQ/PS.

$\sf\large\red{\underline{ }}sin30 = \frac{pq}{20} \\ \\ \frac{1}{2} = \frac{pq}{20}$

∴ PQ = 10 m

$\\ \bf\red{\underline{ }}some \: trigonometric \: identities$

• sin Θ = oposite / hypotenuse
• cosΘ = adjecent / hypotenuse
• tanΘ = opposite/adjecent
• secΘ = hypotenuse / adjecent
• cosecΘ = hypotenuse / opposite
• cotΘ= adjecent / opposite