In Triangle ABC-AB-15cm-AC-8cm- A-50- as shown below- Find the length of BC- -sin 50-0-76-cos 50-0-64-

Drawn CD perpendicular to AB in Right angled triangle ADC nbsp; sin50^∘=CD/AC CD=AC sin50^∘=8× 0.76=6.08 cos 50^∘=AD/AC AD=AC cos50^∘=8× 0.64=5.12 DB= ...

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Byju's Answer

Standard X

Mathematics

Ratio of Sides of a Triangle Determined Using Its Angles

In Triangle A...

Question

In Triangle $A B C, A B = 15 c m, A C = 8 c m, ∠ A = 50^{\circ},$ as shown below. Find the length of BC. $[s i n 50^{\circ} = 0.76, c o s 50^{\circ} = 0.64]$

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Solution

Drawn CD perpendicular to AB

in Right-angled triangle ADC

$s i n 50^{\circ} = \frac{C D}{A C}$

$C D = A C s i n 50^{\circ} = 8 \times 0.76 = 6.08$

$c o s 50^{\circ} = \frac{A D}{A C}$

$A D = A C c o s 50^{\circ} = 8 \times 0.64 = 5.12$

$D B = A B - A D$

$D B = 15 - 5.12 = 9.88 c m$

In right-angled triangle CBD

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

$B C^{2} = (9.88)^{2} + (6.08)^{2}$

$\sqrt{97.61 + 36.966}$

$B C = \sqrt{134.57} = 11.6 c m$

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In Triangle ABC,AB=15cm,AC=8cm,∠A=50∘, as shown below. Find the length of BC. [sin50∘=0.76,cos50∘=0.64]

Drawn CD perpendicular to AB in Right-angled triangle ADC sin50∘=CDAC CD=AC sin50∘=8×0.76=6.08 cos50∘=ADAC AD=AC cos50∘=8×0.64=5.12 DB=AB−AD DB=15−5.12=9.88cm In right-angled triangle CBD BC2=DB2+CD2 BC2=(9.88)2+(6.08)2 √97.61+36.966 BC=√134.57 =11.6cm

In Triangle $A B C, A B = 15 c m, A C = 8 c m, ∠ A = 50^{\circ},$ as shown below. Find the length of BC. $[s i n 50^{\circ} = 0.76, c o s 50^{\circ} = 0.64]$