Advertisements
Advertisements
प्रश्न
In the given figure BD = BC, find the value of x
Advertisements
उत्तर
Given that BD = BC
∆BDC is on isosceles triangle.
In isosceles triangle, angles opposite to equal sides are equal.
∠BDC = ∠BCD ...(1)
Also ∠BCD + ∠BCX = 180° ...[∵ Liner Pair]
∠BCD + 115° = 180°
∠BCD = 180° – 115°
∠BCD = 65° ...[By (1)]
In ∆ADB
∠BAD + ∠ADB = ∠BDC ...[∵ BDC is the exterior angle and ∠BAD and ∠ABD are interior opposite angles]
35° + x = 65°
x = 65° – 35°
x = 30°
APPEARS IN
संबंधित प्रश्न
Find the value of the unknown exterior angle x in the following diagram:

Find the value of the unknown exterior angle x in the following diagram:

Using the measures of the angles given in the figure alongside, find the measures of the remaining three angles.

Find the value of x in the given triangle
Find the value of ‘x’ in the given figure
In the following figure, BC = CA and ∠A = 40°. Then, ∠ACD is equal to ______.

If the exterior angle of a triangle is 130° and its interior opposite angles are equal, then measure of each interior opposite angle is ______.
In ∆PQR, if ∠P = 60°, and ∠Q = 40°, then the exterior angle formed by producing QR is equal to ______.
In ∆ABC, ∠Α = 50°, ∠B = 70° and bisector of ∠C meets AB in D (see figure). Measure of ∠ADC is ______.

A triangle has interior opposite angles measuring 50° and 75°. What is the measure of the exterior angle at the third vertex?

