Subject Test Math 2

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Subject Test Math 2

Question 8 of 28

The figure shows triangle A B C. The length of side A C is labeled x and the length of side B C is labeled 10.

Note: Figure not drawn to scale.

In the figure above, $A B = B C$ and the measure of $∠ C$ is $40^{\circ}$ . What is the value of $x$ ?

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$6.5$

Correct Answer:

No

$6.8$

Correct Answer:

No

$14.7$

Correct Answer:

No

$15.3$

Correct Answer:

Yes

$234.7$

Correct Answer:

No

View Correct Answer

There are several ways to solve this problem. One way is to use the law of sines. Since $A B = B C$ , the measure of $∠ A$ is $40^{\circ}$ and the measure of $∠ B$ is $100^{\circ}$ . Thus, $\frac{10}{\sin 40^{\circ}} = \frac{x}{\sin 100^{\circ}}$ and $x = 15.3$ . (Make sure your calculator is in degree mode.)

You can also use the law of cosines:

$c^{2} = a^{2} + b^{2} - 2 a b \cos C$

$x^{2} = 10^{2} + 10^{2} - 2 (10) (10) \cos 100^{\circ}$

$x^{2} = 234.7296$

$x = 15.3$

Since $Δ A B C$ is isosceles, you can draw the altitude to the triangle.

The figure shows triangle A B C with horizontal side A C and vertex B above side A C. The length of side A B and the length of side B C are both labeled 10. Angle C is labeled 40 degrees. A vertical line segment is drawn from vertex B to side A C. The vertical line segment intersects side A C at a right angle, and the distance between the point of intersection and vertex C is labeled x over 2.

$\cos 40^{\circ} = \frac{\frac{x}{2}}{10}$

$7.66 = \frac{1}{2} x$

$15.32 = x$

Question Difficulty:

medium

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Subject Test Math 2

Question 8 of 28

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