Question

From a triangle of vertices A,B and C of AB 6 cm and AC 10 cm find: a)sinA b)cos A, c)tan C,d)sec C,e)cot A,f)cosec A​

Answer 1

Step-by-step explanation:

by the Pythagoras theorem

AB^2+BC^2=AC^2

BC^2=AC^2-AB^2

BC^2=(10)^2-(6)^2

BC^2=100-36

BC=√(64)

BC=8cm

• sinA=BC/AC=8/10=4/5
• cosA=AB/AC=6/10=3/5
• tanC=AB/BC=6/8=3/4
• secC=AC/BC=10/8=5/4
• cotA=AB/BC=6/8=3/4
• cosecA=AC/BC=10/8=5/4