1條MATHS TRIGO

LET VK be the altitude of ΔVAH
AK=10/2
=5
Thus,VK^2+AK^2=VA^2
=√264

LET VE be the altitude of ΔVBC
BE=10/2
=5

Thus,VE^2+BE^2=VB^2
VE^2+25=196
VE=√171
KE=DC=15

BY COSINE LAW,
15^2=(√264)^2+(√171)^2-2(√264)(√171)(cosKVE)
KVE=60.38399..
KVE=60.4 deg.






√

Draw VH which is thealtitude of ΔVAD, and draw VK which is the altitude of ΔVBC. Then, join HK.

In ΔVAH :
VH² + AH² = VA² (Pythagorean theorem)
VH² + (10/2 cm)² = (17 cm)²
VH = √264 cm

In ΔVBH :
VK² + BH² = VB² (Pythagorean theorem)
VK² + (10/2 cm)² = (14 cm)²
VK = √171 cm

In ΔVHK :
cos∠HVK = [VH² + VK² - HK²] / [2 * VH * VK](cosine law)
cos∠HVK = [(√264)² + (√171)² - 10²] / [2 * √254 x √171]
∠HVK = 38.0°

The angle between ΔVAD and ΔVBC = 38.0°