2 thoughts on “If ΔABC is right angled at C, then the value of Sec (A+B) is ______.”
Answer:
We know that the sum of all three angles of a triangle is always 180°. Given ∠C=90°, hence, ∠A+∠B=180°-90° =90°. We have to find the value of Sec(∠A+∠B)
Step-by-step explanation:
The triangle ABC is given, which is a right-angled triangle with ∠C is equal to 90°.
We know that the sum of all three angles of a triangle is always 180°.
Hence, for ΔABC, ∠A+∠B+∠C =180°
Given ∠C=90°, hence, ∠A+∠B=180°-90° =90°.
We have to find the value of Sec(∠A+∠B).
Hence, Sec(∠A+∠B) = Sec 90° =1/ Cos 90°=1/0=∞. {Since Cos 90°=0}
Answer:
We know that the sum of all three angles of a triangle is always 180°. Given ∠C=90°, hence, ∠A+∠B=180°-90° =90°. We have to find the value of Sec(∠A+∠B)
Step-by-step explanation:
The triangle ABC is given, which is a right-angled triangle with ∠C is equal to 90°.
We know that the sum of all three angles of a triangle is always 180°.
Hence, for ΔABC, ∠A+∠B+∠C =180°
Given ∠C=90°, hence, ∠A+∠B=180°-90° =90°.
We have to find the value of Sec(∠A+∠B).
Hence, Sec(∠A+∠B) = Sec 90° =1/ Cos 90°=1/0=∞. {Since Cos 90°=0}
Therefore, Sec(∠A+∠B)= ∞ (Answer)
Dusra vala math sum nahi pata hai
Answer:
90
Step-by-step explanation:
total sum of angles=180*
90+x+x=180
2x=180-90
2x=90
x=45
so sum of both A and B is 45+45=90