In ∆ ABC ,<C = 90° , BC =21 unit , AB= 29 unit . Find the value of sin A , cos A , sin B and tan B
Answers
Answer:
hope it helps..
Step-by-step explanation:
please make my answer the branliest
Answer:
sinA = 21/29
cosA = 20/29
sinB = 20/29
tanB = 20/21
Step-by-step explanation:
Given ,
ABC forms a triangle
∠C = 90°
BC = 21 units
AB = 29 units
To Find :-
Value of :-
sinA , cosA , sinB , tanB.
How To Do :-
As they said that ∠C = 90° , then ΔABC becomes a right angle triangle at ∠C . So in right angle triangles we can apply the Pythagoras theorem . By applying it we need to find the value of AC and we need find the value of those trigonometric ratios.
Formula Required :-
Pythagoras theorem :-
(Hypotenuse side)² = (opposite side)² + (adjacent side)²
sinα = opposite side/hypotenuse side
cosα = adjacent side/hypotenuse side
tanα = opposite side/adjacent side
Solution :-
Applying Pythagoras theorem :-
(Hypotenuse side)² = (opposite side)² + (adjacent side)²
Considering 'θ' at 'B' :-
∴ Hypotenuse side = AB
Adjacent side = BC
Opposite side = AC
(AB)² = (AC)² + (BC)²
(29)² = (AC)² + (21)²
841 = (AC)² + 441
841 - 441 = (AC)²
400 = (AC)²
AC = √400
→ AC = 20 units
considering 'θ' at ∠A :-
→ Hypotenuse side = AB , opposite side = BC , adjacent side = AC
→ sinA = opposite side/hypotenuse side
= BC/AB
= 21/29
∴ sinA = 21/29
cosA = adjacent side/hypotenuse side
= AC/AB
= 20/29
∴ cosA = 20/29
Considering 'θ' at 'B' :-
Hypotenuse side = AB , Adjacent side = BC , Opposite side = AC
sinB = opposite side/hypotenuse side
= AC/AB
= 20/29
∴ sinB = 20/29
tanB = opposite side/adjacent side
= AC/BC
= 20/21
∴ tanB = 20/21
Note : Refer to the above attachment
