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1 year ago

Basic triangle proofs and congruence only

Mathematics

1 answer:

Sergeu [11.5K]1 year ago

8 0

The Postulate that is used to prove triangle ABC ≅ triangle ADC is SAS Congruence Postulate .

In the question ,

it is given that ,

AD ≅ AB ,

AB is perpendicular to BC , that is AB ⊥ BC

and AD is perpendicular to DC , that is AD ⊥ DC .

Now , in triangle ABC and triangle ADC ,

Side AB = Side AD ......given

angle B = angle D = 90° .....because AB ⊥ BC and AD ⊥ DC .

Side AC = Side AC ......because common side

hence , triangle ABC ≅ triangle ADC by SAS Congruence Postulate .

Therefore , The Postulate that is used to prove triangle ABC ≅ triangle ADC is SAS Congruence Postulate .

Learn more about Congruence here

brainly.com/question/7888063

#SPJ1

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ioda

sin2x =12/13

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STEP - BY - STEP EXPLANATION

What to find?

• sin2x

• cos2x

• tan2x

Given:

tanx = 2/3 = opposite / adjacent

We need to first make a sketch of the given problem.

Let h be the hypotenuse.

We need to find sinx and cos x, but to find sinx and cosx, first determine the value of h.

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From the trigonometric identity;

sin 2x = 2sinxcosx

Substitute the value of sinx , cosx and then simplify.

$\sin 2x=2(\frac{2}{\sqrt[]{13}})(\frac{3}{\sqrt[]{13}})$ $=\frac{12}{13}$

Hence, sin2x = 12/13

cos2x = cos²x - sin²x

Substitute the value of cosx, sinx and simplify.

$\begin{gathered} \cos 2x=(\frac{3}{\sqrt[]{13}})^2-(\frac{2}{\sqrt[]{13}})^2 \\ \\ =\frac{9}{13}-\frac{4}{13} \\ =\frac{5}{13} \end{gathered}$

Hence, cos2x = 5/13

tan2x = 2tanx / 1- tan²x

$\tan 2x=\frac{2\tan x}{1-\tan ^2x}$ $=\frac{2(\frac{2}{3})}{1-(\frac{2}{3})^2}$ $=\frac{\frac{4}{3}}{1-\frac{4}{9}}$ $=\frac{\frac{4}{3}}{\frac{9-4}{9}}$ $=\frac{\frac{4}{3}}{\frac{5}{9}}$ $=\frac{4}{3}\times\frac{9}{5}=\frac{4}{1}\times\frac{3}{5}=\frac{12}{5}$

$\tan 2x=\frac{\sin 2x}{\cos 2x}=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}$

Hence, tan2x = 12/5

Therefore,

sin2x =12/13

cos2x = 5/13

tan2x = 12/5

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