Ashley Ting's answer to Imaboss's Secondary 4 E Maths Singapore question.
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Hihi. Ok im not entirely sure what the ques wants for ii). So if u have the answer sheet, i cld take a look and maybe explain from there. If not, then :)))
Date Posted:
4 years ago
Thanks
Thanks
I presume part b wants something like tan A = 2 tan B or something like that.
From part i,
3 cos A cos B = 5 sin A sin B
"Cross-dividing" cos B and cos A to the RHS ,
3/5 = tan A tan B
so the product of tan A and tan B is 3/5.
From part i,
3 cos A cos B = 5 sin A sin B
"Cross-dividing" cos B and cos A to the RHS ,
3/5 = tan A tan B
so the product of tan A and tan B is 3/5.
Thanks! Any ideas on part (3)
Use the formula which Ashley has written in part ii.
Note that the LHS becomes Tan 45 which equals 1.
Tan A Tan B is found in part ii.
The numerator is what we need to find, by substitution.
Note that the LHS becomes Tan 45 which equals 1.
Tan A Tan B is found in part ii.
The numerator is what we need to find, by substitution.
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