Math - Triangles | Practice Hub

In $\Delta ABC$ , $m \angle A = 66^{\circ }$ , , and . To the nearest tenth, what is $m \angle C$ ?

$39.5^{\circ }$

$39.5^{\circ } \textrm{ or } 140.5^{\circ }$

$73.6^{\circ }$

$73.6^{\circ } \textrm{ or }106.4^{\circ }$

No triangle can exist with these characteristics.

Explanation

Since we are given $m \angle A$ , , and , and want to find $m \angle C$ , we apply the Law of Sines, which states, in part,

$\frac{ \sin m \angle C}{AB} = \frac{ \sin m \angle A}{B C}$ .

Substitute and solve for $\sin m \angle C$ :

$\frac{ \sin m \angle C}{16} = \frac{ \sin 66^{\circ }}{23}$

$\frac{ \sin m \angle C}{16} \cdot 16= \frac{ \sin 66^{\circ }}{23}\cdot 16$

$\sin m \angle C = \frac{ \sin 66^{\circ }}{23}\cdot 16 \approx \frac{ 0.9135}{23}\cdot 16\approx 0.6355$

Take the inverse sine of 0.6355:

$m \angle C = \sin^{-1} 0.6355 \approx 39.5^{\circ }$

There are two angles between $0^{\circ }$ and $180^{\circ }$ that have any given positive sine other than 1 - we get the other by subtracting the previous result from $180^{\circ }$ :

$180 - 39.5 = 140.5^{\circ }$

This, however, is impossible, since this would result in the sum of the triangle measures being greater than $180^{\circ }$ . This leaves $39.5^{\circ }$ as the only possible answer.

Let ABC be a right triangle with sides = 3 inches, = 4 inches, and = 5 inches. In degrees, what is the $\sin \Theta$ where $\Theta$ is the angle opposite of side ?

Explanation

3-4-5_triangle

We are looking for $\sin(\theta )$ . Remember the definition of in a right triangle is the length of the opposite side divided by the length of the hypotenuse.

So therefore, without figuring out $\Theta$ we can find

$\sin \Theta =\frac{3}{5}=.6$

In $\Delta ABC$ , $m \angle A = 66^{\circ }$ , , and . To the nearest tenth, what is $m \angle C$ ?

$39.5^{\circ }$

$39.5^{\circ } \textrm{ or } 140.5^{\circ }$

$73.6^{\circ }$

$73.6^{\circ } \textrm{ or }106.4^{\circ }$

No triangle can exist with these characteristics.

Explanation

Since we are given $m \angle A$ , , and , and want to find $m \angle C$ , we apply the Law of Sines, which states, in part,

$\frac{ \sin m \angle C}{AB} = \frac{ \sin m \angle A}{B C}$ .

Substitute and solve for $\sin m \angle C$ :

$\frac{ \sin m \angle C}{16} = \frac{ \sin 66^{\circ }}{23}$

$\frac{ \sin m \angle C}{16} \cdot 16= \frac{ \sin 66^{\circ }}{23}\cdot 16$

$\sin m \angle C = \frac{ \sin 66^{\circ }}{23}\cdot 16 \approx \frac{ 0.9135}{23}\cdot 16\approx 0.6355$

Take the inverse sine of 0.6355:

$m \angle C = \sin^{-1} 0.6355 \approx 39.5^{\circ }$

There are two angles between $0^{\circ }$ and $180^{\circ }$ that have any given positive sine other than 1 - we get the other by subtracting the previous result from $180^{\circ }$ :

$180 - 39.5 = 140.5^{\circ }$

This, however, is impossible, since this would result in the sum of the triangle measures being greater than $180^{\circ }$ . This leaves $39.5^{\circ }$ as the only possible answer.

In $\Delta ABC$ , $m \angle A = 66^{\circ }$ , , and . To the nearest tenth, what is $m \angle C$ ?

$39.5^{\circ }$

$39.5^{\circ } \textrm{ or } 140.5^{\circ }$

$73.6^{\circ }$

$73.6^{\circ } \textrm{ or }106.4^{\circ }$

No triangle can exist with these characteristics.

Explanation

Since we are given $m \angle A$ , , and , and want to find $m \angle C$ , we apply the Law of Sines, which states, in part,

$\frac{ \sin m \angle C}{AB} = \frac{ \sin m \angle A}{B C}$ .

Substitute and solve for $\sin m \angle C$ :

$\frac{ \sin m \angle C}{16} = \frac{ \sin 66^{\circ }}{23}$

$\frac{ \sin m \angle C}{16} \cdot 16= \frac{ \sin 66^{\circ }}{23}\cdot 16$

$\sin m \angle C = \frac{ \sin 66^{\circ }}{23}\cdot 16 \approx \frac{ 0.9135}{23}\cdot 16\approx 0.6355$

Take the inverse sine of 0.6355:

$m \angle C = \sin^{-1} 0.6355 \approx 39.5^{\circ }$

There are two angles between $0^{\circ }$ and $180^{\circ }$ that have any given positive sine other than 1 - we get the other by subtracting the previous result from $180^{\circ }$ :

$180 - 39.5 = 140.5^{\circ }$

This, however, is impossible, since this would result in the sum of the triangle measures being greater than $180^{\circ }$ . This leaves $39.5^{\circ }$ as the only possible answer.

Let ABC be a right triangle with sides = 3 inches, = 4 inches, and = 5 inches. In degrees, what is the $\sin \Theta$ where $\Theta$ is the angle opposite of side ?

Explanation

3-4-5_triangle

We are looking for $\sin(\theta )$ . Remember the definition of in a right triangle is the length of the opposite side divided by the length of the hypotenuse.

So therefore, without figuring out $\Theta$ we can find

$\sin \Theta =\frac{3}{5}=.6$

Let ABC be a right triangle with sides = 3 inches, = 4 inches, and = 5 inches. In degrees, what is the $\sin \Theta$ where $\Theta$ is the angle opposite of side ?

Explanation

3-4-5_triangle

We are looking for $\sin(\theta )$ . Remember the definition of in a right triangle is the length of the opposite side divided by the length of the hypotenuse.

So therefore, without figuring out $\Theta$ we can find

$\sin \Theta =\frac{3}{5}=.6$

In $\Delta ABC$ , , , and . To the nearest tenth, what is $m \angle A$ ?

$81.3 ^{\circ }$

A triangle with these sidelengths cannot exist.

$81.3 ^{\circ }\textrm{ or } 98.7^{ \circ }$

$98.7^{ \circ }$

$171.3^{ \circ }$

Explanation

By the Triangle Inequality, this triangle can exist, since .

By the Law of Cosines:

$\left ( BC\right )^{2} = \left ( AB\right )^{2} + \left ( AC\right )^{2} - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

Substitute the sidelengths and solve for $\cos m \angle A$ :

$32^{2} = 26^{2} + 23^{2} - 2\cdot 26 \cdot 23 \cdot \cos m \angle A$

$1,024 =676 + 529 - 1,196 \cdot \cos m \angle A$

$1,024 =1,205 - 1,196 \cdot \cos m \angle A$

$-181 = - 1,196 \cdot \cos m \angle A$

$\cos m \angle A = 181 \div 1,196 \approx 0.1513$

$m \angle A \approx \cos^{-1} 0.1513 \approx 81.3 ^{\circ }$

In $\Delta ABC$ , $m \angle A = 75^{\circ }$ , $m \angle B = 62^{\circ }$ , and . To the nearest tenth, what is ?

Explanation

Since we are given and want to find , we apply the Law of Sines, which states, in part,

$\frac{ \sin m \angle A}{B C} = \frac{ \sin m \angle C}{AB}$

$m \angle A = 75^{\circ }$ and

$m \angle C = 180 - \left ( m \angle A + m \angle B \right )= 180 -(75+ 62) = 180 -137 = 43^{\circ }$

Substitute in the above equation:

$\frac{ \sin75^{\circ }}{16} = \frac{ \sin 43^{\circ }}{AB}$

Cross-multiply and solve for :

$AB\cdot \sin75^{\circ } = 16 \cdot \sin 43^{\circ }$

$AB = \frac{16 \cdot \sin 43^{\circ }}{\sin75^{\circ }} = \frac{16 \cdot \0.6820}{0.9659} \approx 11.3$

In $\Delta ABC$ , $m \angle A = 75^{\circ }$ , $m \angle B = 62^{\circ }$ , and . To the nearest tenth, what is ?

Explanation

Since we are given and want to find , we apply the Law of Sines, which states, in part,

$\frac{ \sin m \angle A}{B C} = \frac{ \sin m \angle C}{AB}$

$m \angle A = 75^{\circ }$ and

$m \angle C = 180 - \left ( m \angle A + m \angle B \right )= 180 -(75+ 62) = 180 -137 = 43^{\circ }$

Substitute in the above equation:

$\frac{ \sin75^{\circ }}{16} = \frac{ \sin 43^{\circ }}{AB}$

Cross-multiply and solve for :

$AB\cdot \sin75^{\circ } = 16 \cdot \sin 43^{\circ }$

$AB = \frac{16 \cdot \sin 43^{\circ }}{\sin75^{\circ }} = \frac{16 \cdot \0.6820}{0.9659} \approx 11.3$

In $\Delta ABC$ , , , and . To the nearest tenth, what is $m \angle A$ ?

$81.3 ^{\circ }$

A triangle with these sidelengths cannot exist.

$81.3 ^{\circ }\textrm{ or } 98.7^{ \circ }$

$98.7^{ \circ }$

$171.3^{ \circ }$

Explanation

By the Triangle Inequality, this triangle can exist, since .

By the Law of Cosines:

$\left ( BC\right )^{2} = \left ( AB\right )^{2} + \left ( AC\right )^{2} - 2 \cdot AB \cdot AC \cdot \cos m \angle A$

Substitute the sidelengths and solve for $\cos m \angle A$ :

$32^{2} = 26^{2} + 23^{2} - 2\cdot 26 \cdot 23 \cdot \cos m \angle A$

$1,024 =676 + 529 - 1,196 \cdot \cos m \angle A$

$1,024 =1,205 - 1,196 \cdot \cos m \angle A$

$-181 = - 1,196 \cdot \cos m \angle A$

$\cos m \angle A = 181 \div 1,196 \approx 0.1513$

$m \angle A \approx \cos^{-1} 0.1513 \approx 81.3 ^{\circ }$

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