Question

17. In Fig. 16.13, ABC is an equilateral triangle of side
6 cm. D is a point in BC such that BD = 1 cm and E
is the mid-point of BC. Calculate :
(i) AE,
(ii) tan angleADC.
(iii) angleADC to the nearest degree,
(iv) angleBAD to the nearest degree.

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Answer 1

Answer:

(1) as its an equilateral triangle , AE is perpendicular to BC

Therefore AE = $\sqrt{6^{2} - 3^{2} }$ = 5 cm

(2) angle ADC= angle ADE

therefore from triangle ADE, tan ADC  = AE/DE = 5/2

(3) tan inverse of 5/2 = Angle ADC = 68.18 degree

(4) angle DAE = 180 - ( angle ADC + angle AED) = 180 - (68 + 90) = 22 degree

Anglr BAE = 180 - (angle AEB +  angle ABE) = 180 - (90+60) = 30

therefore angle BAD= angle BAE - angle DAE = 30-22 = 8 degrees

Step-by-step explanation: