If ∠A and ∠B are acute angles such that cos A = cos B we have proved that ∠A = ∠B since AC = BC and angles opposite to equal sides of a triangle are equal.
Step-by-step explanation:
Using the basic trigonometric ratios, we can solve this problem.
In the right-angled triangle ABC as shown below, ∠A and ∠B are acute angles and ∠C is right angle.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
cos A = side adjacent to ∠A / hypotenuse = AC/AB
cos B = side adjacent to ∠B / hypotenuse = BC/AB
Given that cos A = cos B
Therefore, AC/AB = BC/AB
AC = BC
Hence, ∠A = ∠B (angles opposite to equal sides of a triangle are equal.)
Let's look into an alternative approach to solve the question.
Let us consider a triangle ABC in which CO ⊥ AB.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
It is given that cos A = cos B
AO/AC = BO/BC
AO/BO = AC/BC
Let AO/BO = AC/BC = k
AO = k × BO ...(i)
AC = k × BC ...(ii)
By applying Pythagoras theorem in ΔCAO and ΔCBO, we get
If ∠A and ∠B are acute angles such that cos A = cos B we have proved that ∠A = ∠B since AC = BC and angles opposite to equal sides of a triangle are equal.
Answers & Comments
Answer:
If ∠A and ∠B are acute angles such that cos A = cos B we have proved that ∠A = ∠B since AC = BC and angles opposite to equal sides of a triangle are equal.
Step-by-step explanation:
Using the basic trigonometric ratios, we can solve this problem.
In the right-angled triangle ABC as shown below, ∠A and ∠B are acute angles and ∠C is right angle.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
cos A = side adjacent to ∠A / hypotenuse = AC/AB
cos B = side adjacent to ∠B / hypotenuse = BC/AB
Given that cos A = cos B
Therefore, AC/AB = BC/AB
AC = BC
Hence, ∠A = ∠B (angles opposite to equal sides of a triangle are equal.)
Let's look into an alternative approach to solve the question.
Let us consider a triangle ABC in which CO ⊥ AB.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
It is given that cos A = cos B
AO/AC = BO/BC
AO/BO = AC/BC
Let AO/BO = AC/BC = k
AO = k × BO ...(i)
AC = k × BC ...(ii)
By applying Pythagoras theorem in ΔCAO and ΔCBO, we get
AC2 = AO2 + CO2 (from ΔCAO)
CO2 = AC2 - AO2 ...(iii)
BC2 = BO2 + CO2 (from ΔCBO)
CO2 = BC2 - BO2 ...(iv)
From equation (iii) and equation (iv), we get
AC2 - AO2 = BC2 - BO2
(kBC)2 - (kBO)2 = BC2 - BO2 [From equation (i) and (ii)]
k2BC2 - k2BO2 = BC2 - BO2
k2 (BC2 - BO2) = BC2 - BO2
k2 = (BC2 - BO2)/(BC2 - BO2) = 1
k = 1
Putting this value in equation (ii) we obtain,
AC = BC
Thus, ∠A = ∠B (angles opposite to equal sides of triangle are equal.)
Verified answer
Answer:
If ∠A and ∠B are acute angles such that cos A = cos B we have proved that ∠A = ∠B since AC = BC and angles opposite to equal sides of a triangle are equal.
hope it helps you:)
hello hru?