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In △ABC, m∠A=53 °, m∠B=17 °, and a=27 . Find the perimeter of the triangle. law of sines 4

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Answer:

perimeter for given △ABC equals approximately 75.974 round to up to 76 units.

Explanation:

The perimeter of the triangle in question can be determined by using the Law of Sines.

According to this law, the following equation holds true---

a/sinA=b/sinB=c/sinC, where a is side opposite to angle A, b is side opposite angle B and c is side opposite to angle C. In our △ABC, m∠A = 53° and m∠B = 17°, so if we substitute these values into the equation together with known side length a = 27 , we get:

27/sin53°=b/ sin17° → b= 27*(sin17)/(Sin53)= 14.8228... ≈15.

So now that we know all three sides (a = 27 , b = 15 and c), it's easy to calculate the perimeter of our triangle: P = a + b + c; P=27+15+c ;P = 42 + c .

The last missing piece -the length of side "c"- can be obtained by using Pythagoras theorem on right-angled triangle consisting of sides "a" and "b".

We know that a2+b2=c2→ 282+(152)= c2→ 902+225=1125 → √1125≈ 33 .5834 ... ≈ 34

So finally, plugging in this value for "c" into our perimeter formula yields: P = 42 + 34; P ≈ 76

Therefore, the perimeter for given △ABC equals approximately 76 units.

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