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love history [14]
3 years ago
9

Some isosceles triangles are not equilateral. a true or b false

Mathematics
1 answer:
Bas_tet [7]3 years ago
8 0
That's true.  The definition of an isosceles triangle is that it has at least 2 sides equal in length.  Therefore, isosceles can be equilateral, but equilateral cannot be isosceles.
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Express tan D as a fraction in simplest terms. 40 B 24 32 C​
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\sf Tan \ D = \dfrac{4}{3}

Step-by-step explanation:

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       \sf \boxed{\bf \tan \ D =\dfrac{opposite \ side \ of \ \angle \ D}{adjacent \ side \ of \ \angle D}}

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3 years ago
Given △DEF, which is not equal to cos(F)? sin(F). sin(D). tan(F). cos(D)
valentina_108 [34]

Answer: The correct option is tan(F).

Explanation:

In the given figure the triangle is a right angle triangle because \angle E=90^{\circ}.

It is also given that DE=EF=5 and DF=5\sqrt{2}.

Since it is an isosceles right angle triangle, therefore the value of perpendicular and base is same for both angles D and F, which is 5.

\cos (F)=\frac{Base}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}

The value of cos(F) is \frac{1}{\sqrt{2}}.

\sin (F)=\frac{Perpendicular}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}

The value of sin(F) is \frac{1}{\sqrt{2}}.

\sin (D)=\frac{Perpendicular}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}

The value of sin(D) is \frac{1}{\sqrt{2}}.

\tan (F)=\frac{Perpendicular}{Base}=\frac{5}{5} =1

The value of tan(F) is 1. Which is not equal to \frac{1}{\sqrt{2}}.

\cos (D)=\frac{Base}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}

The value of cos(D) is \frac{1}{\sqrt{2}}.

Therefore, the value of tan(F) is not equal to the value of cos(F), so the correct option is third.

3 0
3 years ago
Read 2 more answers
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