In a - ABC- if - C -30- a -47 cm and b -94 cm- then the triangle isA- Right angledB- Obtuse angledC- Right angled isoscelesD- Isosceles

The correct option is B Obtuse angled cosC = a^2 + b^2 c^2/2ab ⇒√(3)/2 = (47)^2 + (94)^2 c^2/2 × 47 × 94⇒ c = 58.24 ∠ B = 124^∘. Hence obtuse angled triang ...

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Standard XII

Mathematics

Cosine Rule

In a Δ ABC, i...

Question

In a $Δ A B C, i f ∠ C = 30^{\circ}, a = 47 c m a n d b = 94 c m$ , then the triangle is

Right angled

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Right angled isosceles

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Isosceles

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Obtuse angled

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Solution

The correct option is D
Obtuse angled

$c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$
$\Rightarrow \frac{\sqrt{3}}{2} = \frac{(47)^{2} + (94)^{2} - c^{2}}{2 \times 47 \times 94} \Rightarrow c = 58.24$
$∠ B = 124^{\circ}$ . Hence obtuse angled triangle.

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In a $Δ A B C, i f ∠ C = 30^{\circ}, a = 47 c m a n d b = 94 c m$ , then the triangle is

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In a ΔABC,if∠C=30∘,a=47cm and b=94cm, then the triangle is

The correct option is D Obtuse angled cosC=a2+b2−c22ab ⇒√32=(47)2+(94)2−c22×47×94⇒c=58.24 ∠B=124∘. Hence obtuse angled triangle.

In a $Δ A B C, i f ∠ C = 30^{\circ}, a = 47 c m a n d b = 94 c m$ , then the triangle is

The correct option is D
Obtuse angled

$c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$
$\Rightarrow \frac{\sqrt{3}}{2} = \frac{(47)^{2} + (94)^{2} - c^{2}}{2 \times 47 \times 94} \Rightarrow c = 58.24$
$∠ B = 124^{\circ}$ . Hence obtuse angled triangle.