In triangle $abc$, $m/angle a = 90^/circ$, $m/angle b = 75^/circ$, and $bc = / sqrt {3}$ units. what is the area of triangle $abc$? express your answer as a common fraction. your answer submittips give up

Question

James Bailey

Mathematics

13 October, 01:30

In triangle $abc$, $m/angle a = 90^/circ$, $m/angle b = 75^/circ$, and $bc = / sqrt {3}$ units. what is the area of triangle $abc$? express your answer as a common fraction. your answer submittips give up

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Answers (1)

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Leah Wells · Answer 1

As we can see from the angles given, one is 90 degrees, the triangle should be a right triangle. We are given the two angles and one side which would represent the hypotenuse or the longest side. The area of a triangle is calculated from one half the product of its base and its height. From the given values, we need to calculate for the height and base. We do as follows:

sin 75 = height / √3

height = 1.67

cos 75 = base / √3

base = 0.45

Area = bh/2

Area = (0.45) (1.67) / 2

Area = 0.37 square units

Therefore, the area of the triangle is about 0.37 square units.