How do you solve the triangle when sides of a = 6, b = 10 and measure of angle A = (in degrees, minutes, seconds) 31˚10', find the other sides (c) and other angle measures (B)(C)?

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How do you solve the triangle when sides of a = 6, b = 10 and measure of angle A = (in degrees, minutes, seconds) 31˚10', find the other sides (c) and other angle measures (B)(C)?

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Answer 1 · 2017-02-25T07:41:01

$c=11.593$ , $/_B=59.695^@$ and $/_C=89.228^@$

Explanation:

To find the third side given two other sides $a=6$ and $b=10$ and included angle $/_A=31^@10'=31.167^@$ , as $10'$ is one-sixth of $1^@$ ). Now using sine formula

$a/sinA=b/sinB=c/sinC$ i.e.

$6/(sin31.167^@)=10/sinB=c/sinC$

or $6/0.5175=10/sinB=c/sinC$

Hence $sinB=(10xx0.5175)/6=0.8626$ and $/_B=59.695^@$

and $/_C=180^@-31.167^@-59.695^@=89.228^@$

and $c=6/0.5175xxsin89.228^@=6/0.5175xx0.9999=11.593$

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How do you solve the triangle when sides of a = 6, b = 10 and measure of angle A = (in degrees, minutes, seconds) 31˚10', find the other sides (c) and other angle measures (B)(C)?

1 Answer

Explanation:

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