in the given figure angle ACB = 90 ,BDC=90, CD = 4CM,BD = 3 CM , AC = 12 CM, - Maths - Introduction to Trigonometry

Question

in the given figure angle ACB = 90 ,BDC=90, CD = 4CM,BD = 3 CM , AC
= 12 CM, find the value of cosA- sin A ?

Vijay Kumar Gupta · Accepted Answer

In △BDC, ∠D=90°Use pythagoras to first find the length of BC     BC2=BD2+CD2            =32+42             =9+16            =25This gives,  BC2=25   BC=5Now in △ABC, ∠ACB=90°Use pythagoras to first find the length of AB     AB2=AC2+BC2            =122+52             =144+25            =169This gives,  AB2=169   AB=13Note that sine of an angle is the ratio of the length of perpendicluar to the length of hypotenuse
and the cosine of an angle is the ratio of the length of base to the length of hypotenuseThis gives,   sinA=BCAB=513   cosA=ACAB=1213So the required value is,   cosA-sinA=1213-513                        =12-513                        =713

in the given figure angle ACB = 90 ,BDC=90, CD = 4CM,BD = 3 CM , AC = 12 CM, find the value of cosA- sin A ?