Math, asked by pratibha08sr303, 3 months ago

in a triangle ABC ,AD perpendicular to BC ,AB=7.5cm,AC=10 cm AD=6cm . show that angle BAC=90⁰​

Answers

Answered by vandan92
5

Answer:

The value of BC=12.5\ cmBC=12.5 cm and \angle BAC=90\ degree∠BAC=90 degree in \triangle ABC△ABC .

Step-by-step explanation:

Given,

In \triangle ABC△ABC where AD\perp BC,AB=7.5\ cm,AC=10\ cm\ and\ AD=6\ cmAD⊥BC,AB=7.5 cm,AC=10 cm and AD=6 cm

From figure,

\angle ADB=\angle ADC=90\ degree∠ADB=∠ADC=90 degree

In \triangle ABD△ABD ,

AB^{2}=AD^{2}+BD^{2}AB

2

=AD

2

+BD

2

⇒AB^{2}-AD^{2}=BD^{2}AB

2

−AD

2

=BD

2

⇒(7.5)^{2}-6^{2}=BD^{2}(7.5)

2

−6

2

=BD

2

⇒BD=\sqrt{56.25-36}BD=

56.25−36

⇒BD=\sqrt{20.25}BD=

20.25

⇒BD=4.5\ cmBD=4.5 cm

In \triangle ACD△ACD ,

CD=\sqrt{AC^{2}-AD^{2} }CD=

AC

2

−AD

2

⇒CD=\sqrt{10^{2}-6^{2} }CD=

10

2

−6

2

⇒DC=8\ cmDC=8 cm

∴ BC=BD+CD=4.5+8=12.5\ cmBC=BD+CD=4.5+8=12.5 cm

From \triangle ABC△ABC ,

BC=\sqrt{AB^{2}+AC^{2} }BC=

AB

2

+AC

2

⇒BC=\sqrt{7.5^{2}+10^{2} }BC=

7.5

2

+10

2

⇒BC=12.5\ cmBC=12.5 cm

So, The value of BC=12.5\ cmBC=12.5 cm and \angle BAC=90\ degree∠BAC=90 degree in \triangle ABC△ABC .

Step-by-step explanation:

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