How do you solve a triangle given a=10 A=20 degrees c=11?

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How do you solve a triangle given a=10 A=20 degrees c=11?

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Answer 1 · 2018-07-31T16:36:58

( i ) B = 137.9^o, C = 22.1^p and b = 19.6, nearly.
( ii ) B = 2.1^o, C = 157.9^p and b = 1.07, nearly.

Explanation:

Use $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} and A + B + C = 180^{o}$ $a/sin A = b/sin B = c/sin C and A + B + C = 180^o$ .

$sin C = (\frac{c}{a}) sin A = 1.1 {sin 20}^{o}$ $sin C =(c/a) sin A = 1.1 sin 20^o$

$\Rightarrow C = arcsin (1.1 {sin 20}^{o}) = {22.1}^{o} and 180^{o} - {22.1}^{o}$ $rArr C = arcsin( 1.1 sin 20^o ) = 22.1^o and 180^o - 22.1^o$

$= {157.9}^{o}$ $= 157.9^o$ , nearly.#

$B = 180^{o} - A - C = {137.9}^{o}$ $B = 180^o - A - C = 137.9^o$ or 2.1^o#, nealy.,

nearly.

$b = a (\frac{sin B}{sin A})$ $b = a ( sin B/sin A)$

$= 10 \frac{sin ({137.9}^{o})}{{sin 20}^{o}} and 10 \frac{sin {{2.1}^{o}}}{{sin 20}^{o} =}$ $= 10 sin ( 137.9^o )/(sin 20^o) and 10 sin {2.1^o}/sin 20^o =$

$= 19.60 and 1.07$ $= 19.60 and 1.07$ , nearly.

Note: 10 = BC = a is a chord of a circle subtending 20^o at A, on

the circle. There are two positions, at which c = BA is 11.

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How do you solve a triangle given a=10 A=20 degrees c=11?

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