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erik [133]
3 years ago
11

Given two angles that measure 50 degrees and 80 degrees and side that measures 4 feet, how many triangles, if any, can be constr

ucted?
Mathematics
1 answer:
kirza4 [7]3 years ago
3 0
The angles are the only constraint here that counts.  If one of the three interior angles of a supposed triangle is 50 degrees and another is 80 degrees, then the third angle must be 50 degrees.  Thus, we have a 50-50-80 triangle, which is isosceles though not a right triangle.  If 4 feet is a measure of one of the equal sides of a supposed triangle, then obviously the adjacent side also has measure 4 ft.

The set of angles remains the same (50-50-80), but subject to the constraint mentioned above, the measure of any one of the sides has infinitely many possible values, so long as those values are positive.

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Question 8: Find the equation of the straight line that:
Scorpion4ik [409]

Answer:

y=4x+6

Step-by-step explanation:

Hi there!

Linear equations are typically organized in slope-intercept form: y=mx+b where m is the slope (also called the gradient) and b is the y-intercept (the value of y when x is 0)

<u>1) Plug the gradient into the equation (b)</u>

y=mx+b

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y=4x+b

<u>2) Determine the y-intercept (b)</u>

y=4x+b

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Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.
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Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ

Likewise:

\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

m\angle PAC=m\angle PAB+m\angle BAC

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m\angle QAB=m\angle QAC+m\angle BAC

Since ∠QAC ≅ ∠PAB:

m\angle PAC=m\angle QAC+m\angle BAC

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