Use the law of Sines to find the missing angle of the triangle. Find m
1 answer:
Ah...Trigonometry is fun!
The law of sines states:
The transitive property (switching the orders of the equations) applies here. Therfore, we can say that
We then plug in our given values to find C
Solving, we get 0.8557316387. We're not done yet!
We are trying to find an angle measure, so we'll do the inverse of the ratio we used (sin).
arcsin0.8557316387 (arcsin is the same as inverse sin)
=58.8 (approximate)
So the measure of angle C is 58.8. You could check this by reinserting it into the equation.
:)
The law of sines states:
The transitive property (switching the orders of the equations) applies here. Therfore, we can say that
We then plug in our given values to find C
Solving, we get 0.8557316387. We're not done yet!
We are trying to find an angle measure, so we'll do the inverse of the ratio we used (sin).
arcsin0.8557316387 (arcsin is the same as inverse sin)
=58.8 (approximate)
So the measure of angle C is 58.8. You could check this by reinserting it into the equation
:)
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Hello!
We have two probabilities we can use; we have 170/400, for our experiment, and 1/2, which is our theoretical probability.
To solve, we just multiply the two probabilities.
=0.2125≈21.3
Therefore, we have about a 21.3% chance of this event occurring.
I hope this helps!
Your question can be quite confusing, but I think the gist of the question when paraphrased is: P<span>rove that the perpendiculars drawn from any point within the angle are equal if it lies on the angle bisector?
Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like </span>∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector
Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like </span>∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector