In - ABC-AB-10 cm- AC-6cm- - A-70- Find the length of the side BC- -cos 70-0-34- sin 70-0-94-

Draw nbsp;CD nbsp;perpendicular to nbsp; AB. In right angle triangle ACD sin 70^∘ =CD/AC CD=AC.sin 70^∘ CD=6× 0.94 = 5.64 cm cos 70^∘ =AD/AC AD=A ...

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Byju's Answer

Standard X

Mathematics

Ratio of Sides of Special Triangles

In Δ ABC,AB=1...

Question

In $Δ A B C, A B = 10 c m, A C = 6 c m, ∠ A = 70^{\circ}$ . Find the length of the side BC. $[c o s 70^{\circ} = 0.34, s i n 70^{\circ} = 0.94]$

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Solution

Draw CD perpendicular to AB.

In right angle triangle ACD

$s i n 70^{\circ} = \frac{C D}{A C}$

$C D = A C . s i n 70^{\circ}$

$C D = 6 \times 0.94 = 5.64 c m$

$c o s 70^{\circ} = \frac{A D}{A C}$

$A D = A C c o s 70^{\circ}$

$A D = 6 \times 0.34 = 2.04 c m$

$B D = A B - A D = (10 - 2.04) = 7.96 c m$

In right angled triangle BCD

$B C^{2} = B D^{2} + C D^{2}$

$B C = \sqrt{(7.96)^{2} + (5.64)^{2}}$

$B C = \sqrt{31.81 + 63.36}$

$B C = \sqrt{95.17} = 9.75 c m .$

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In ΔABC,AB=10cm,AC=6cm,∠A=70∘. Find the length of the side BC. [cos 70∘=0.34,sin 70∘=0.94]

Draw CD perpendicular to AB. In right angle triangle ACD sin70∘=CDAC CD=AC.sin 70∘ CD=6×0.94 = 5.64 cm cos 70∘=ADAC AD=AC cos 70∘ AD=6×0.34 = 2.04 cm BD=AB−AD=(10−2.04)=7.96cm In right angled triangle BCD BC2=BD2+CD2 BC=√(7.96)2+(5.64)2 BC=√31.81+63.36 BC=√95.17 =9.75cm.

In $Δ A B C, A B = 10 c m, A C = 6 c m, ∠ A = 70^{\circ}$ . Find the length of the side BC. $[c o s 70^{\circ} = 0.34, s i n 70^{\circ} = 0.94]$