Find the numerical value of x from the diagram given below.

In Δ ABC, such that AB = BC. And BC = BD. AB is extended till point E such that ∠EBD = 75°. Let ∠BAC = x°. If two sides of a triangle are equal, then two opposite angles are always equal. ∠BAC = ∠BCA = x° (∴ AB = BC) Sum of all angles in triangle ABC is 180°. ⇒ ∠BAC + ∠BCA + ∠CBA = 180° ⇒ x° + x° + ∠CBA = 180° ⇒ 2x° + ∠CBA = 180° ⇒ ∠CBA = 180° - 2x° Similarly, in Δ BDC, Let ∠CBD = θ If two sides of a triangle are equal, then two opposite angles are always equal. ∠CBD = ∠CDB = θ (∴ DB = BC) Sum of all angles in triangle BDC is 180°. ⇒ ∠CBD + ∠CDB + ∠DCB = 180° ⇒ θ + θ + ∠DCB = 180° ⇒ 2θ + ∠DCB = 180° ⇒ ∠DCB = 180° - 2θ ∠DCB and ∠BCA form linear angle. ⇒ 180° - 2θ + x = 180° ⇒ - 2θ + x = 0 ⇒ 2θ = x ∠EBD, ∠DBC and ∠CBA form linear angle. ⇒ ∠EBD + ∠DBC + ∠CBA = 180° ⇒ 75° + θ + 180° - 2x° = 180° (∵ Substituting x = 2θ) ⇒ 75° + θ + 180° - 2(2θ) = 180° ⇒ 75° + θ - 4θ = 0 ⇒ 75° - 3θ = 0 ⇒ 3θ = 75° ⇒ θ = ⇒ θ = 25° So, x = 2θ = 2 x 25° = 50° Hence, the value of x = 50°.

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If a = 1 + log 2 - log 5, b = 2 log 3, c = log P - log 5 and a + b = 2c, find the value of P.

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In △PQR; PQ = PR. A is a point in PQ and B is a point in PR, so that QR = RA = AB = BP.
(i) Show that: ∠P : ∠R = 1 : 3
(ii) Find the value of ∠Q.

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The given figure shows a right triangle right angled at B.
If ∠BCA = 2∠BAC, show that AC = 2BC.

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Mathematics

Find the numerical value of x from the diagram given below.

Triangles

Related Questions

If a = 1 + log 2 - log 5, b = 2 log 3, c = log P - log 5 and a + b = 2c, find the value of P.

In △PQR; PQ = PR. A is a point in PQ and B is a point in PR, so that QR = RA = AB = BP.
(i) Show that: ∠P : ∠R = 1 : 3
(ii) Find the value of ∠Q.

The given figure shows a right triangle right angled at B.
If ∠BCA = 2∠BAC, show that AC = 2BC.

Mathematics

Find the numerical value of x from the diagram given below.

Triangles

Related Questions

If a = 1 + log 2 - log 5, b = 2 log 3, c = log P - log 5 and a + b = 2c, find the value of P.

In △PQR; PQ = PR. A is a point in PQ and B is a point in PR, so that QR = RA = AB = BP.(i) Show that: ∠P : ∠R = 1 : 3(ii) Find the value of ∠Q.

The given figure shows a right triangle right angled at B.If ∠BCA = 2∠BAC, show that AC = 2BC.

In △PQR; PQ = PR. A is a point in PQ and B is a point in PR, so that QR = RA = AB = BP.
(i) Show that: ∠P : ∠R = 1 : 3
(ii) Find the value of ∠Q.

The given figure shows a right triangle right angled at B.
If ∠BCA = 2∠BAC, show that AC = 2BC.