Q3 of 25 Page 276

In ΔABC, AC=BC and BC is extended upto the point D. If ∠ACD=144°, then let us determine the circular value of each of the angles of ΔABC.


∠ACD = 144°


∵ DCB is straight line, ∠DCB = 180°


⇒ ∠ACB = 180° - 144° = 36°


∵ AC=BC, by opposite angle property


⇒ ∠CAB = ∠CBA = x


By Angle sum Property


⇒ ∠CAB + ∠CBA + ∠ACB = 180°


⇒ 2x + 36° = 180°


⇒ x = 72°


⇒ ∠CAB = ∠CBA = 72° and ∠ACB = 36°




And ∠ACB = 36°




More from this chapter

All 25 →
1

Let us express the following into degrees, minutes and seconds.

(i) 832’ (ii) 6312’’


(iii) 375’’ (iv)


(v) 72.04°

2

Let us determine the circular values of the followings

(i) 60° (ii) 135°


(ii) -150° (iv) 72°


(v) 22°30’ (vi) -62°30’


(vii) 52°52’30’’

 4

If the difference of two acute angles of a right-angled triangle is , then let us write the sexagesimal values of two angles.

 5

The measure of one angle of a triangle is 65° and other angle is ; let us write the sexagesimal value and circular value of third angle.