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Lelechka [254]
2 years ago
11

Pls answer for brainlest and 50 points PLEASE EXPLAIN In image

Mathematics
2 answers:
anzhelika [568]2 years ago
8 0
Answer: C) 163

Step-by-Step Solution:

In the Right Triangle formed to the extreme right, lets mark the angles as
∠1, ∠2 and ∠3.

Therefore, from the Figure :-

∠1 = 73°
∠2 = 90°

By Angle Sum Property :-

∠3 = 180 - (73 + 90)
∠3 = 180 - 163
=> ∠3 = 17°

The Angle which forms a Linear Pair with ∠3 is the Corresponding Angle of ∠r, and Corresponding Angles are Equal.

Therefore,
=> 180 - ∠3
= 180 - 17
=> 163°

Therefore, the Angle that forms the Linear Pair with ∠3 is 163°

This Angle is Corresponding to ∠r and hence they are Equal ie. ∠r = 163°

Hence, ∠r = 163°
AlexFokin [52]2 years ago
5 0

First, let's complete the angles in the triangle. Remember that the sum of the angles in a triangle is 180 degrees.

73 + 90 + x = 180

163 + x = 180

x = 17

So, the angle that completes the triangle is 17 degrees. If we look at that angle in the triangle and the one adjacent to it, we can see that those two angles form a linear pair (or are supplementary, both meaning that they add up to 180 degrees).

17 + x = 180

x = 163

So, 17's supplement is 163 degrees. The 163 degree angle corresponds with angle r, and corresponding angles are congruent.

Therefore, angle r is 163 degrees. The correct answer is option C.

Hope this helps!

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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6 0
4 years ago
"In quadrilateral QRST, m∠Q is 68°, m∠R is (3x + 40)°, and m∠T is (5x − 52)°. What are the measures of ∠R , ∠S , and ∠T ? Write
34kurt
TQRS is an inscribed quadrilateral.
5 x - 52° + 3 x + 40° = 180°
8 x - 12° = 180°
8 x = 180° + 12°
8 x = 192°
x = 192° : 8 = 24°
m∠ R = 3 · 24° + 40° = 112°
m∠ T = 5 · 24° - 52° = 68°
m∠ S = 360° - ( 68° + 68° + 112° ) = 112°
Answer:  
m∠R, m∠S, m∠T = 112°, 112°, 68°.
4 0
3 years ago
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There is actually 2 questions
Zanzabum

Answer:

not  sure but i need points

Step-by-step explanation:

6 0
3 years ago
To find the measure of ∠A in ∆ABC, use the___(Pythagorean Theorem, Law of Sines, Law of Cosines). To find the length of side HI
nadya68 [22]

<u>Part 1) </u>To find the measure of ∠A in ∆ABC, use

we know that

In the triangle ABC

Applying the law of sines

\frac{a}{sin\ A}=\frac{b}{sin\ B}=\frac{c}{sin\ C}

in this problem we have

\frac{a}{sin\ A}=\frac{b}{sin\ theta}\\ \\a*sin\ theta=b*sin\ A\\ \\ sin\ A=\frac{a*sin\ theta}{b} \\ \\ A=arc\ sin (\frac{a*sin\ theta}{b})

therefore

<u>the answer  Part 1) is</u>

Law of Sines

<u>Part 2) </u>To find the length of side HI in ∆HIG, use

we know that

In the triangle HIG

Applying the law of cosines

g^{2}=h^{2}+i^{2}-2*h*i*cos\ G

In this problem we have

g=HI

G=angle Beta

substitute

HI^{2}=h^{2}+i^{2}-2*h*i*cos\ Beta

HI=\sqrt{h^{2}+i^{2}-2*h*i*cos\ Beta}

therefore

<u>the answer Part 2) is</u>

Law of Cosines

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