We are given a tasked to find the geometric property of a triangle that satisfies the given conditionm∠abc=m∠cbd, then m∠cbd=m∠abc.

It can be observed that if the first value is equal to the second value then the second value is also equal to the first value. This kind of characteristics describes the Symmetric Property.

3 0

3 years ago

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If a^n=b than a is the n^th root of b. True or false

dem82 [27]

Step-by-step explanation:

yes that's true

if a^n=b

then

a^n/n=b^1/n

a=b^1/n

7 0

3 years ago

Find the values of the sine, cosine, and tangent for ZA.<br><br> (TOP OF TRIANGLE IS (A))

Sloan [31]

$\bigstar\:{\underline{\sf{In\:right\:angled\:triangle\:ABC\::}}}\\\\$

AC = 7 m
BC = 4 m

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$\bf{\dag}\:{\underline{\frak{By\:using\:Pythagoras\: Theorem,}}}\\\\$

$\star\:{\underline{\boxed{\frak{\purple{(Hypotenus)^2 = (Perpendicular)^2 + (Base)^2}}}}}\\\\\\ :\implies\sf (AB)^2 = (AC)^2 + (BC)^2\\\\\\ :\implies\sf (AB)^2 = (AB)^2 = (7)^2 = (4)^2\\\\\\ :\implies\sf (AB)^2 = 49 + 16\\\\\\ :\implies\sf (AB)^2 = 65\\\\\\ :\implies{\underline{\boxed{\pmb{\frak{AB = \sqrt{65}}}}}}\:\bigstar\\\\$

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☆ Now Let's find value of sin A, cos A and tan A,

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sin A = Perpendicular/Hypotenus = $\sf \dfrac{4}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{4 \sqrt{65}}{65}}$

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cos A = Base/Hypotenus = $\sf \dfrac{7}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{7 \sqrt{65}}{65}}$

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tan A = Perpendicular/Base = ${\sf{\pink{\dfrac{4}{7}}}}$

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$\therefore\:{\underline{\sf{Hence,\: {\pmb{Option\:A)}}\:{\sf{is\:correct}}.}}}$

4 0

2 years ago

If the endpoints of AB have the coordinates A(9, 8) and B(-1, -2), what is the midpoint of AB?

pickupchik [31]

The midpoint, if the endpoints are (x₁,y₁) and (x₂,y₂):
$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

$(9,8) \\
x_1=9 \\ y_1=8 \\ \\
(-1,-2) \\
x_2=-1 \\ y_2=-2 \\ \\
\frac{x_1+x_2}{2}=\frac{9-1}{2}=\frac{8}{2}=4 \\
\frac{y_1+y_2}{2}=\frac{8-2}{2}=\frac{6}{2}=3$

The midpoint of AB is (4,3).

5 0

2 years ago

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