Question 1

LEVEL 6

A triangle ABC is shown below:

Angle x is an obtuse angle.

Find the value of x.

Select the correct answer from the list below:

A: 123.4\degree

B: 108.5\degree

C: 115.8\degree

D: 125.9\degree

 

CORRECT ANSWER:  A: 123.4\degree

WORKED SOLUTION:

To calculate x we can use the cosine rule,

\text{cos}(A)=\dfrac{b^2+c^2-a^2}{2bc}

Hence substituting the values in,

\text{cos}(x)=\dfrac{110^2+100^2-185^2}{2(110)(100)}

Taking the inverse cosine of both sides,

x= \text{cos}^{-1} \bigg( \dfrac{-12125}{22000} \bigg) =123.4\degree

Question 2

LEVEL 6

A triangle ABC is shown below:

Angle x is an acute angle.

Find the value of x.

Select the correct answer from the list below:

A: 58.3 \degree

B: 64.9 \degree

C: 55.7 \degree

D: 61.1 \degree

 

CORRECT ANSWER:  B: 61.1 \degree

WORKED SOLUTION:

To calculate x we can use the cosine rule,

\text{cos}(A)=\dfrac{b^2+c^2-a^2}{2bc}

Hence substituting the values in,

\text{cos}(x)=\dfrac{11^2+8^2-10^2}{2(11)(8)}

Taking the inverse cosine of both sides,

x=\text{cos}^{-1} \bigg( \dfrac{85}{176} \bigg) =61.1\degree

Question 3

LEVEL 6

A triangle ABC is shown below:

Find the value of x.

Select the correct answer from the list below:

A: 5.62 cm

B: 14.42 cm

C: 24.51 cm

D: 4.95 cm

 

CORRECT ANSWER:  D: 4.95 cm

WORKED SOLUTION:

To calculate x, we can use the cosine rule,

a^2 = b^2 + c^2 - 2bc \, \text{cos}(A)

Hence substituting the values in,

x^2 = 8^2 + 12^2 - 2 \times 8 \times 12 \times \text{cos}(35) = 24.5099...

Taking the square root of both sides, and only taking the positive value of x, since x represents a real physical length, we get

x = \sqrt{24.5099...} = 4.95 cm

Question 4

LEVEL 6

A triangle ABC is shown below:

Find the value of x.

Select the correct answer from the list below:

A: 41.36 cm

B: 36.06 cm

C: 29.83 cm

D: 45.56 cm

 

CORRECT ANSWER:  A: 41.36 cm

WORKED SOLUTION:

To calculate x, we can use the cosine rule,

a^2 = b^2 + c^2 - 2bc \, \text{cos}(A)

Hence substituting the values in,

x^2 = 20^2 + 30^2 - 2 \times 20 \times 30 \times \text{cos}(110) = 1710.42...

Taking the square root of both sides, and only taking the positive value of x, since x represents a real physical length, we get

x = \sqrt{1710.42...} = 41.36 cm

Question 5

LEVEL 6

A triangle ABC is shown below:

Angle x is an acute angle.

Find the value of x.

Select the correct answer from the list below:

A: 85.66 \degree

B: 90.34 \degree

C: 93.82 \degree

D: 82.55 \degree

 

CORRECT ANSWER:  C: 93.82 \degree

WORKED SOLUTION:

To calculate x we can use the cosine rule,

\text{cos}(A)=\dfrac{b^2+c^2-a^2}{2bc}

Hence substituting the values in,

\text{cos}(x)=\dfrac{3^2+5^2-6^2}{2(3)(5)}

Taking the inverse cosine of both sides,

x= \text{cos}^{-1} \bigg( \dfrac{-2}{30} \bigg) = 93.82 \degree