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# In the Given Figure. Find Rp and Ps Using the Information Given in ∆Psr. - Geometry

Concept30 - 60 - 90 and 45 - 45 - 90 Theorem

#### Question

In the given figure. Find RP and PS using the information given in ∆PSR.

#### Solution

In ∆PSR,
∠S = 90, ∠P = 30, ∴ ∠R = 60
By 30∘ − 60 − 90 theorem,

$\text{RS} = \frac{1}{2} \times \text{RP}$
$\Rightarrow 6 = \frac{1}{2} \times \text{RP}$
$\Rightarrow 6 \times 2 = \text{RP}$
$\Rightarrow \text{RP} = 12 . . . \left( 1 \right)$
$\text{PS} = \frac{\sqrt{3}}{2} \times \text{RP}$
$= \frac{\sqrt{3}}{2} \times 12$
$= 6\sqrt{3} . . . \left( 2 \right)$
Hence, RP = 12 and PS = 6$\sqrt{3}$
Is there an error in this question or solution?

#### APPEARS IN

Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 2: Pythagoras Theorem
Practice Set 2.1 | Q: 4 | Page no. 39

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Solution In the Given Figure. Find Rp and Ps Using the Information Given in ∆Psr. Concept: 30 - 60 - 90 and 45 - 45 - 90 Theorem.
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