In a triangle , let∠PQR=30°and the sides PQ and QR have lengths 103 and 10, respectively. Then , which of the following statement(s) is (are) TRUE?

cos30°=(103)2+(10)2-(PR)22×103×10⇒PR=10∵QR=PR⇒∠PQR=∠QPR ∠QPR=30° Area of ΔPQR=12×103×10×sin30°=12×10×103×12∣=253∠QRP=180°-30°+30°=120°r=ΔS=25310+10+1032=25310+53=53·(2-3)=103-15R=a2sinA=102sin30°=10 ∴ Area =πR2=100π

In a triangle , let∠PQR=30°and the sides PQ and QR have lengths 103 and 10, respectively. Then , which of the following statement(s) is (are) TRUE?

cos30°=(103)2+(10)2-(PR)22×103×10⇒PR=10∵QR=PR⇒∠PQR=∠QPR ∠QPR=30° Area of ΔPQR=12×103×10×sin30°=12×10×103×12∣=253∠QRP=180°-30°+30°=120°r=ΔS=25310+10+1032=25310+53=53·(2-3)=103-15R=a2sinA=102sin30°=10 ∴ Area =πR2=100π

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In a triangle 𝑃𝑄𝑅, $l e t ∠ P Q R = 30 °$ and the sides PQ and QR have lengths $10 \sqrt{3}$ and 10, respectively. Then , which of the following statement(s) is (are) TRUE?

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$∠ Q P R = 45 °$

$The area of the triangle PQR is 25 \sqrt{3} and ∠ Q R P = 120 °$

$The radius of the incircle of the triangle PQR is 10 \sqrt{3} - 15$

2,3,4 $The area of the circumcircle of the triangle P Q R is 100 π$

answer is B, C, D.

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Detailed Solution

$\cos 30 ° = \frac{(10 \sqrt{3})^{2} + (10)^{2} - (P R)^{2}}{2 \times 10 \sqrt{3} \times 10}$

$\begin{array}{l} \Rightarrow P R = 10 ∵ Q R = P R \Rightarrow ∠ P Q R = ∠ Q P R ∠ Q P R = 30 ° \\ Area of Δ P Q R = \frac{1}{2} \times 10 \sqrt{3} \times 10 \times \sin 30 ° = \frac{1}{2} \times 10 \times 10 \sqrt{3} \times \frac{1}{2} ∣ \\ = 25 \sqrt{3} \end{array}$

$\begin{array}{l} ∠ Q R P = 180 ° - (30 ° + 30 °) = 120 ° \\ r = \frac{Δ}{S} = \frac{25 \sqrt{3}}{(\frac{10 + 10 + 10 \sqrt{3}}{2})} = \frac{25 \sqrt{3}}{10 + 5 \sqrt{3}} = 5 \sqrt{3} \cdot (2 - \sqrt{3}) = 10 \sqrt{3} - 15 \\ R = \frac{a}{2 \sin A} = \frac{10}{2 \sin 30 °} = 10 ∴ Area = π R^{2} = 100 π \end{array}$