Question #b03a6

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Question #b03a6

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Answer 1 · 2017-10-09T20:01:59

$27^{o}$ $27^o$ To nearest degree.

Explanation:

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Since it doesn't specify which side is the hypotenuse, the triangle will be marked in the conventional manner, with side c being the hypotenuse.

Because of the large values we will reduce the dimensions by a factor of one hundred. This will not affect the size of the relevant angles .i.e similar triangles.

$c^{2} = a^{2} + b^{2}$ $c^2 = a^2+b^2$

$\to c^{2} = {(4.9)}^{2} + {(9.6)}^{2} = 116.17 \Rightarrow c = \sqrt{116.17}$ $-> c^2=(4.9)^2+(9.6)^2=116.17=> c=sqrt(116.17)$

Since we now know all three sides and one angle, we can use the Sine Rule:

$\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ $sinA/a=sinB/b=sinC/c$

We are looking for angle A and we know angle C. So:

$\frac{sin A}{a} = \frac{sin C}{c} \Rightarrow sin A = \frac{a sin C}{c}$ $sinA/a= sinC/c=> sinA=(asinC)/c$

$\to sin A = \frac{(4.9) sin (90)}{\sqrt{116.17}} = \frac{4.9}{\sqrt{116.17}}$ $-> sinA= ((4.9)sin(90))/(sqrt(116.17))=4.9/(sqrt(116.17))$

$A = arcsin (\frac{4.9}{\sqrt{116.17}}) = {27.04}^{o}$ $A= arcsin((4.9)/(sqrt(116.17)))=27.04^o$ ( 2.d.p.)

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Question #b03a6

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