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a) In a triangle AABC with side lengths a, b, c opposite to the angles A,B,C respectively, prove the cosine law: a² +c² – b² cos B = 2ас h b) If D is the foot from C, show that 4s(s – a)(s – b)(s – b) B A |CD]? = h² = c² a + b +c where s 2 (Hint: Note that s – a =(b +c - a) etc. ) c) Prove that the area of a triangle AABC is given by JABC| = /s(s – a)(s – b)(s – c),

a) In a triangle AABC with side lengths a, b, c opposite to the angles A,B,C respectively, prove the cosine law: a² +c² – b² cos B = 2ас h b) If D is the foot from C, show that 4s(s – a)(s – b)(s – b) B A |CD]? = h² = c² a + b +c where s 2 (Hint: Note that s – a =(b +c - a) etc. ) c) Prove that the area of a triangle AABC is given by JABC| = /s(s – a)(s – b)(s – c),

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter8: Areas Of Polygons And Circles
Section8.5: More Area Relationships In The Circle
Problem 9E
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a) In a triangle AABC with side lengths a, b, c opposite to the angles A,B,C respectively, prove the cosine law: a² +c² – b² cos B = 2ас h b) If D is the foot from C, show that 4s(s – a)(s – b)(s – b) B A |CD]? = h² = c² a + b +c where s 2 (Hint: Note that s – a =(b +c - a) etc. ) c) Prove that the area of a triangle AABC is given by JABC| = /s(s – a)(s – b)(s – c),
Transcribed Image Text:a) In a triangle AABC with side lengths a, b, c opposite to the angles A,B,C respectively, prove the cosine law: a² +c² – b² cos B = 2ас h b) If D is the foot from C, show that 4s(s – a)(s – b)(s – b) B A |CD]? = h² = c² a + b +c where s 2 (Hint: Note that s – a =(b +c - a) etc. ) c) Prove that the area of a triangle AABC is given by JABC| = /s(s – a)(s – b)(s – c),
Expert Solution
Step 1

Given- In a triangle ABC with sides a,b, c opposite to the angles A,B,C respectively.

Algebra homework question answer, step 1, image 1

To prove- The cosine Law ,cosB=a2+c2-b22ac

Step 2

Explanation- As per the figure, in the △BCD,we can write as,

CosB=BDa

So, we can write as,

      BD=acosB            ⋯⋯1

Now, using pythagoras theorem in △ACD,we get,

AC2=CD2+DA2

Further we can write as,

b2=h2+c-BD2

Now, from the equation (1), sustituting the values of BD from the equation (1), we get,

b2=h2+c-acosB2b2=h2+c2+a2cos2B-2accosB               ⋯⋯2

Now, from the figure, we can write as, 

h2+a2cos2B=a2

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