Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

# The length of the sides of a triangle are 8, 8, and 3. What is the measure of the smallest angle in the triangle?

A)20.6

B)21.6

C)22.0

D)28.4

E)33.8

* There is clearly something wrong with the way i solved this, i used the Law of Cosines, the answer i got os none of these five. It would be great if someone explained. Thanks:)

Relevance

Simple mistakes are made by many when using the cosine rule. However, you need to show your workings so we can identify where your mistakes are being made.

Anyway, the smallest angle is opposite the shortest side so,

3² = 8² + 8² - 2(8)(8)cosθ

so,  128cosθ = 8² + 8² - 3²

i.e.  128cosθ = 119

Hence, cosθ = 119/128

so, θ = cos⁻¹(119/128) => 21.6°

:)>

• Use the Cosine Rule

3^2 = 8^2 + 8^2 - 2(8)(8)CosA

9 = 64 + 64 - 128CosA

CosA = (9 - 64 - 64 ) / -128

CosA = -119/-128

CosA = 119/128

CosA = 0.92968...

A = Cos^-1( 0.92968...)

A = 21.613845... degrees.

• 8x8x3 , none its cant be that size .

• Since the triangle is isosceles, A/2 = sin⁻¹(1.5/8) = 10.8º. The smallest angle A = 2*10.8º = 21.6º

• Triangle with Three sides Known (SSS) -

The Angles are:  79.19 /  79.19 and 21.6

Use the Law of Cosines First to find one of the Angles. Then use the Law of Cosines again to find another Angle.  The last angle is found by 180 - A - B.

COS A = (8^2 + 8^2 - 3^2) / 2*8*8

COS A = (64 + 64 - 9) / 128

COS A = 0.9297

A = COS^-1 (0.9297)

A = 21.6138

• Let the small angle be A, then

cosA=(8^2+8^2-3^2)/(2*8^2)

=>

cosA=1-9/(2*64)=0.9296875

=>

A=cos^-1(0.9296875)=21.6 (B)

•  The length of the sides of a triangle are 8, 8, and 3.

c^2 = a^2 + b^2 − 2ab cos (C)

9 = 64 + 64 − 2(64) cos (C)

cos (C) = 119/128

C = 34.2675° (degrees)

B or C = 21.465° (degrees)

The measure of the smallest angle in the triangle:

21.465° (degrees)

• Visualize 2 8'' arms of an isosceles triangle descending from the apex to opposite ends of a 3'' base. Label the apex A and base BC. Sides are labeled as a,b and c where a = BC, b = AC and c = AB. By law of cosines we have

a^2 = b^2+c^2-2bc*cosA. Then cosA = (b^2+c^2-a^2)/2bc. Here a = 3, b=c=8.

cosA=(8^2+8^2-3^2)/128=119/128 so A = arccos(119/128) = 21.61384575 deg

The smallest side is opposite the smallest angle and visa-versa. That's why I

chose the apex angle as the smallest angle in the triangle. Correct answer is B).

• Anonymous
2 months ago

Shortest edge has smallest angle.

angle A = acos((b.b+c.c-a.a)/(2bc)), Law of Cosines

= acos((8.8.2-3.3)/(2.8.8)