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