Right Triangles and Similar Triangles - SAT Math

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Question

Fire tower A is miles due west of fire tower B. Fire tower A sees a fire in the direction degrees west of north. Fire tower B sees the same fire in the direction degrees east of north. Which tower is closer to the fire and by how much?

Answer

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle. This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry.

$\sin 50^\circ = \frac{AF}{2}$

$AF \approx 1.53 miles$

$\sin 40^\circ = \frac{BF}{2}$

$BF \approx 1.29 miles$

Fire tower B is miles closer to the fire.

Compare your answer with the correct one above

Question

Find the length of side .

Answer

In an angle-side-angle problem, Law of Sines will solve the triangle.

First find angle A:

Then use Law of Sines.

$\frac {AB}{\sin 83^\circ} = \frac {4}{\sin 53^\circ} \rightarrow AB = \frac {4 \sin 83^\circ}{\sin 53^\circ} \approx 5.0$

Compare your answer with the correct one above

Question

Find the area of the triangle.

Answer

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

$A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18$

Compare your answer with the correct one above

Question

What is the length of the leg of a right triangle whose hypotenuse is 5cm and other leg is 4cm?

Answer

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c).

Subtracting 16 from either side of the equation gives us:

The last step is to take the square root both sides resulting in:

Compare your answer with the correct one above

Question

Fire tower A is miles due west of fire tower B. Fire tower A sees a fire in the direction degrees west of north. Fire tower B sees the same fire in the direction degrees east of north. Which tower is closer to the fire and by how much?

Answer

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle. This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry.

$\sin 50^\circ = \frac{AF}{2}$

$AF \approx 1.53 miles$

$\sin 40^\circ = \frac{BF}{2}$

$BF \approx 1.29 miles$

Fire tower B is miles closer to the fire.

Compare your answer with the correct one above

Question

Find the length of side .

Answer

In an angle-side-angle problem, Law of Sines will solve the triangle.

First find angle A:

Then use Law of Sines.

$\frac {AB}{\sin 83^\circ} = \frac {4}{\sin 53^\circ} \rightarrow AB = \frac {4 \sin 83^\circ}{\sin 53^\circ} \approx 5.0$

Compare your answer with the correct one above

Question

Find the area of the triangle.

Answer

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

$A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18$

Compare your answer with the correct one above

Question

What is the length of the leg of a right triangle whose hypotenuse is 5cm and other leg is 4cm?

Answer

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c).

Subtracting 16 from either side of the equation gives us:

The last step is to take the square root both sides resulting in:

Compare your answer with the correct one above

Question

Fire tower A is miles due west of fire tower B. Fire tower A sees a fire in the direction degrees west of north. Fire tower B sees the same fire in the direction degrees east of north. Which tower is closer to the fire and by how much?

Answer

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle. This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry.

$\sin 50^\circ = \frac{AF}{2}$

$AF \approx 1.53 miles$

$\sin 40^\circ = \frac{BF}{2}$

$BF \approx 1.29 miles$

Fire tower B is miles closer to the fire.

Compare your answer with the correct one above

Question

Find the length of side .

Answer

In an angle-side-angle problem, Law of Sines will solve the triangle.

First find angle A:

Then use Law of Sines.

$\frac {AB}{\sin 83^\circ} = \frac {4}{\sin 53^\circ} \rightarrow AB = \frac {4 \sin 83^\circ}{\sin 53^\circ} \approx 5.0$

Compare your answer with the correct one above

Question

Find the area of the triangle.

Answer

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

$A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18$

Compare your answer with the correct one above

Question

What is the length of the leg of a right triangle whose hypotenuse is 5cm and other leg is 4cm?

Answer

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c).

Subtracting 16 from either side of the equation gives us:

The last step is to take the square root both sides resulting in:

Compare your answer with the correct one above

Question

Fire tower A is miles due west of fire tower B. Fire tower A sees a fire in the direction degrees west of north. Fire tower B sees the same fire in the direction degrees east of north. Which tower is closer to the fire and by how much?

Answer

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle. This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry.

$\sin 50^\circ = \frac{AF}{2}$

$AF \approx 1.53 miles$

$\sin 40^\circ = \frac{BF}{2}$

$BF \approx 1.29 miles$

Fire tower B is miles closer to the fire.

Compare your answer with the correct one above

Question

Find the length of side .

Answer

In an angle-side-angle problem, Law of Sines will solve the triangle.

First find angle A:

Then use Law of Sines.

$\frac {AB}{\sin 83^\circ} = \frac {4}{\sin 53^\circ} \rightarrow AB = \frac {4 \sin 83^\circ}{\sin 53^\circ} \approx 5.0$

Compare your answer with the correct one above

Question

Find the area of the triangle.

Answer

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

$A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18$

Compare your answer with the correct one above

Question

What is the length of the leg of a right triangle whose hypotenuse is 5cm and other leg is 4cm?

Answer

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c).

Subtracting 16 from either side of the equation gives us:

The last step is to take the square root both sides resulting in:

Compare your answer with the correct one above

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Right Triangles and Similar Triangles - SAT Math

Fire tower A is miles due west of fire tower B. Fire tower A sees a fire in the direction degrees west of north. Fire tower B sees the same fire in the direction degrees east of north. Which tower is closer to the fire and by how much?

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle. This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry. Fire tower B is miles closer to the fire.

Find the length of side .

In an angle-side-angle problem, Law of Sines will solve the triangle. First find angle A: Then use Law of Sines.

Find the area of the triangle.

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

What is the length of the leg of a right triangle whose hypotenuse is 5cm and other leg is 4cm?

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c). Subtracting 16 from either side of the equation gives us: The last step is to take the square root both sides resulting in:

Fire tower A is miles due west of fire tower B. Fire tower A sees a fire in the direction degrees west of north. Fire tower B sees the same fire in the direction degrees east of north. Which tower is closer to the fire and by how much?

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle. This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry. Fire tower B is miles closer to the fire.

Find the length of side .

In an angle-side-angle problem, Law of Sines will solve the triangle. First find angle A: Then use Law of Sines.

Find the area of the triangle.

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

What is the length of the leg of a right triangle whose hypotenuse is 5cm and other leg is 4cm?

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c). Subtracting 16 from either side of the equation gives us: The last step is to take the square root both sides resulting in:

Fire tower A is miles due west of fire tower B. Fire tower A sees a fire in the direction degrees west of north. Fire tower B sees the same fire in the direction degrees east of north. Which tower is closer to the fire and by how much?

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle. This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry. Fire tower B is miles closer to the fire.

Find the length of side .

In an angle-side-angle problem, Law of Sines will solve the triangle. First find angle A: Then use Law of Sines.

Find the area of the triangle.

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

What is the length of the leg of a right triangle whose hypotenuse is 5cm and other leg is 4cm?

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c). Subtracting 16 from either side of the equation gives us: The last step is to take the square root both sides resulting in:

Fire tower A is miles due west of fire tower B. Fire tower A sees a fire in the direction degrees west of north. Fire tower B sees the same fire in the direction degrees east of north. Which tower is closer to the fire and by how much?

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle. This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry. Fire tower B is miles closer to the fire.

Find the length of side .

In an angle-side-angle problem, Law of Sines will solve the triangle. First find angle A: Then use Law of Sines.

Find the area of the triangle.

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

What is the length of the leg of a right triangle whose hypotenuse is 5cm and other leg is 4cm?

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c). Subtracting 16 from either side of the equation gives us: The last step is to take the square root both sides resulting in:

In an angle-side-angle problem, Law of Sines will solve the triangle.

First find angle A:

Then use Law of Sines.

$\frac {AB}{\sin 83^\circ} = \frac {4}{\sin 53^\circ} \rightarrow AB = \frac {4 \sin 83^\circ}{\sin 53^\circ} \approx 5.0$

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

$A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18$

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c).

Subtracting 16 from either side of the equation gives us:

The last step is to take the square root both sides resulting in:

In an angle-side-angle problem, Law of Sines will solve the triangle.

First find angle A:

Then use Law of Sines.

$\frac {AB}{\sin 83^\circ} = \frac {4}{\sin 53^\circ} \rightarrow AB = \frac {4 \sin 83^\circ}{\sin 53^\circ} \approx 5.0$

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

$A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18$

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c).

Subtracting 16 from either side of the equation gives us:

The last step is to take the square root both sides resulting in:

In an angle-side-angle problem, Law of Sines will solve the triangle.

First find angle A:

Then use Law of Sines.

$\frac {AB}{\sin 83^\circ} = \frac {4}{\sin 53^\circ} \rightarrow AB = \frac {4 \sin 83^\circ}{\sin 53^\circ} \approx 5.0$

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

$A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18$

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c).

Subtracting 16 from either side of the equation gives us:

The last step is to take the square root both sides resulting in:

In an angle-side-angle problem, Law of Sines will solve the triangle.

First find angle A:

Then use Law of Sines.

$\frac {AB}{\sin 83^\circ} = \frac {4}{\sin 53^\circ} \rightarrow AB = \frac {4 \sin 83^\circ}{\sin 53^\circ} \approx 5.0$

Dropping the altitude creates two special right triangles as shown in the diagram. Use the area formula of a triangle to get

$A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18$

One leg is 4cm and the hypotenuse is 5cm. Plug in 4 for one of the legs and 5 for the hypotenuse (c).

Subtracting 16 from either side of the equation gives us:

The last step is to take the square root both sides resulting in: