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barxatty [35]
3 years ago
5

Which step can be used to prove that triangle EFG is also a right triangle?

Mathematics
1 answer:
Darya [45]3 years ago
8 0
EFG is the same as angle KLM. therefore, it is the reflection of the angle KLM
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Ira Lisetskai [31]

Answer:

∠MLP = 72° ,           ∠LJK = 22° ,            ∠JKL = 72° ,          ∠KLJ  =  86°

Step-by-step explanation:

Here, given In ΔJLK and  ΔMLP

Here,  JK  II  ML,  LM = MP

∠JLM = 22° and  ∠LMP = 36°

Now, As angles opposite to equal sides are equal.

⇒ ∠MLP = ∠MPL  = x°

Now, in  ΔMLP

By <u>ANGLE SUM PROPERTY</u>:   ∠MLP + ∠MPL  + ∠LMP = 180°

⇒ x° + x° + 36° = 180°

⇒ 2 x  = 180 - 36 = 144

or, x  = 72°

⇒ ∠MLP = ∠MPL  = 72°

Now,as  JK  II  ML

⇒ ∠LJK = ∠JLM = 22° ( Alternate pair of angles)

Now, by the measure of straight angle:

∠MLP + ∠JLM + ∠JLK = 180°  ( Straight angle)

⇒ 72° + 22° + ∠JLK = 180°

or, ∠JLK  =  86°

In , in  ΔJLK

By <u>ANGLE SUM PROPERTY</u>:   ∠JKL + ∠JLK  + ∠LJK = 180°

⇒  ∠JKL + 86° + 22° = 180°

⇒ ∠JKL   = 180 - 108 = 72 , or ∠JKL = 72°

Hence, from  above proof ,  ∠MLP = 72° , ∠LJK = 22° , ∠JKL = 72° ,

∠KLJ  =  86°

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