# Pre-calc questions? So many triangles….it’s confusing?

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1.)

a = 12, b = 22, C = 95°

``````a. c ≈ 26, A ≈ 27.6°, B ≈ 57.4°

b. c ≈ 26, A ≈ 54.4°, B ≈ 30.6°

c. c ≈ 25.1, A ≈ 31.6°, B ≈ 53.4°   ``````

2.)

Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary.

B = 105°, a = 12 cm, c = 19 cm

``````a. 114 cm2

b. 29.51 cm2

c. 110.12 cm2   ``````

3.)

Determine whether a triangle can be formed with the given side lengths. If so, use Heron’s formula to find the area of the triangle.

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a = 240

b = 122

c = 177

``````a. 10,414.95

b. 5883.43

c. No triangle is formed.``````

• if we assume that angle A is opposite side a, we have two sides and the angle between them which means we use the law of cosines to find the missing side. from there we can use the law of sines to find the other angles in question 1. so (c^2) = (a^2) + (b^2) – 2ab(cosC) ; (c^2) =(12^2) + (22^2) – [(2)(12)(22)cos95] ;

c = sqrt(674.018) = 25.961 so now we can eliminate choice c so it is between choice a and choice b. so now using law of sines [ 25.961 / sin95] = [12 / sinA] cross multiply and solve for sinA ;

25.961(sinA) = 12sin95 ; sinA = [12sin95]/25.961 = .46047 ; take inverse sine of both sides to solve for A;

A = 27.4 degrees closest choice is answer A so 1) is A . you would use the law of cosines for number 2 to find the length of side b and then use Herons formula to find the area of the triangle. for number 3, if the sides form a triangle then a+b > c AND a+c > b AND b+c > a ; so 362 > 177 , 417 > 122 , and 299 > 240 so yes these sides do form a triangle so we use Herons formula to find the area ; s = [a+b+c] / 2 = 269.5

so Area = sqrt( 269.5(269.5 – 240)(269.5 – 177)(269.5 – 122) ) = 10414.95 which is answerA 