If △ABC△ABC△ABC is right angled at CCC, find the value of cos(A+B)cos(A+B)cos(A+B).
Answer:
000
- We know that sum of the angles of a triangle is 180∘180∘180∘
∠A+∠B+∠C=180∘∠A+∠B+∠C=180∘∠A+∠B+∠C=180∘ - It is given that angle CCC is right angle. Replace ∠C=90∘∠C=90∘∠C=90∘ in above equation
⟹∠A+∠B+90∘=180∘⟹∠A+∠B+90∘=180∘⟹∠A+∠B+90∘=180∘
⟹∠A+∠B=180∘−90∘⟹∠A+∠B=180∘−90∘⟹∠A+∠B=180∘−90∘
⟹∠A+∠B=90∘⟹∠A+∠B=90∘⟹∠A+∠B=90∘
- Now we know the value of A+BA+BA+B, we can find the value of cos(A+B)cos(A+B)cos(A+B)
⟹cos(A+B)=cos(90∘)⟹cos(A+B)=cos(90∘)⟹cos(A+B)=cos(90∘)
⟹cos(A+B)=0⟹cos(A+B)=0⟹cos(A+B)=0