In triangle ABC - - CAB -40- AC -6 cm - AB -7 cm - The length BC is equal to- -sin 40-0-67- cos 40-0-76-A- 6-55 mB- 4-55 mC- 5-65 mD- 7-25 m

The correct option is B 4.55 m Draw CD⊥ nbsp; nbsp;AB nbsp;In ADC sin 40^∘ = CD/AC 0.64= CD/6 CD = 6 × 0.64 = 3.85 m cos 40^∘= AD/AC 0.76 = AD/6 AD ...

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Standard X

Mathematics

Sine Rule

In triangle A...

Question

In triangle ABC, $∠ C A B = 40^{\circ}$ , AC = 6 cm, AB = 7 cm. The length BC is equal to_____. $[s i n 40^{\circ} = 0.67, c o s 40^{\circ} = 0.76]$

7.25 m

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6.55 m

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4.55 m

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5.65 m

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Solution

The correct option is C
4.55 m

Draw CD $⊥ A B$

In $△ A D C sin 40^{\circ} = \frac{C D}{A C} 0.64 = \frac{C D}{6} C D = 6 \times 0.64 = 3.85 m cos 40^{\circ} = \frac{A D}{A C} 0.76 = \frac{A D}{6} A D = 6 \times 0.76 A D = 4.56 m From figure, we can see that D B = A B - A D D B = 7 - 4.56 D B = 2.44 m In right-angled triangle CDB$

$B C^{2} = B D^{2} + D C^{2} B C^{2} = {(2.44)}^{2} + {(3.85)}^{2} B C^{2} = 5.95 + + 17.82 B C^{2} = 20.77 B C = 4.55 m$

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In triangle ABC, ∠CAB=40∘, AC = 6 cm, AB = 7 cm. The length BC is equal to_____. [sin 40∘=0.67, cos 40∘=0.76]

The correct option is C 4.55 m Draw CD⊥AB In △ADCsin40∘=CDAC0.64=CD6CD=6×0.64=3.85 mcos40∘=ADAC0.76=AD6AD=6×0.76AD=4.56 mFrom figure, we can see thatDB=AB−ADDB=7−4.56DB=2.44 mIn right-angled triangle CDB BC2=BD2+DC2BC2=(2.44)2+(3.85)2BC2=5.95++17.82BC2=20.77BC=4.55 m

In triangle ABC, $∠ C A B = 40^{\circ}$ , AC = 6 cm, AB = 7 cm. The length BC is equal to_____. $[s i n 40^{\circ} = 0.67, c o s 40^{\circ} = 0.76]$