Hypotenuse and angle ratio relationshipIn triangle ∠ B A C = 90, ∠ A B C: A C B = 1 : 2 and A C = 4 c m. Calculate the length of BC. I am interested in the solution using Pythagoras theorem.

Question

Hypotenuse and angle ratio relationshipIn triangle   ∠  B  A  C  =  90,   ∠  A  B  C:       A    C    B    =  1  :  2 and   A  C  =  4  c  m. Calculate the length of BC. I am interested in the solution using Pythagoras theorem.

gortepap6yb · Accepted Answer

Step 1Well if   ∠  B  A  C  =      90    ∘   and   ∠  A  B  C  :  ∠  A  C  B  =  1  :  2, this means   ∠  A  B  C  =      60    ∘   and   ∠  A  C  B  =      30    ∘  .Step 2Now   A  C  =  4 and   B  C  =      A          B      2        +    A          C      2      . But   A  B  =  B  C  cos  ⁡  (      60    ∘    ).

anraszbx · Accepted Answer

Step 1  ∠  B  A  C  =      90    o    ,  ∠  A  B  C  :  ∠  A  C  B  =  1  :  2  ,  A  C  =  4Draw a line through A such that   ∠  B  A  D  =      60    o  . Let it intersect BC at D. Now ADB is an equilateral triangle. So:  A  D  =  D  B  =  B  A  ,  ∠  A  D  B  =  ∠  D  B  A  =  ∠  B  A  D  =      60    o  As   ∠  A  D  C  =      180    0    −  ∠  A  D  B  =      120    o  In   △  A  C  D  ∠  D  A  C  =      180    o    −  ∠  A  C  D  −  ∠  C  D  A  =      180    o    −      30    o    −      120    o    =      30    o  So   ∠  A  C  D  =  ∠  C  A  D, so sides   A  D  =  C  D.Now let   B  A  =  x, so   A  B  =  B  D  =  D  A  =  x    A  D  =  C  D  =  xStep 2Apply pythagoreas theorem in   △  A  B  C:  A      B    2    +  A      C    2    =  B      C    2  So:       x    2    +  A      C    2    =  (  2  x      )    2    =  4      x    2      A      C    2    =  3      x    2      ⟹    A  C  =      3    xAs:   A  C  =      3    x  =  4    ⟹    x  =  4      /        3  Hence                               B          C          =          2          x          =          8                      /                                3

In triangle ABC angle BAC=90, angle ABC: angle ACB=1:2 and AC = 4cm. Calculate the length of BC.

Answered question

Answer & Explanation

New Questions in High school geometry